Control some function by the average of its extreme values on an interval

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Let $lambda geq 1$ and $varepsilon leq frac12sqrtlambda$.

Why is it the case that $exists C$ a universal constant (i.e. that does not depend on $lambda, varepsilon$) such that $forall i in mathbbN, forall t in [lambda - varepsilon , lambda + varepsilon ]$,
$$e^-t t^i leq frac12 C left( e^-(lambda - varepsilon)(lambda - varepsilon)^i + e^-(lambda + varepsilon)(lambda + varepsilon)^i right) $$

The careful reader will have noticed that this is a property about the pmf of the Poisson distribution. For information, this statement appears in "An Automatic Inequality Prover and Instance Optimal Identity testing -- G. Valiant & P. Valiant, Lemma 7" , but it isn't as obvious to me as it is to them.

Here is were we are at so far:

Write $f_i(t) = e^-tt^i$ for simplicity of notation, we then want to show that there exists $C$ universal such that $f_i(t) leq frac12 C left( f_i(lambda - varepsilon) + f_i(lambda + varepsilon) right) $.

Notice that since $f_i'(t) = f_(i-1)(t)(i-t)$, $f_i$ is increasing on $[0, i]$ and decreasing on $[i, infty]$.

For $i < lambda - varepsilon$ it is the case that $f_i(t) leq f_i(lambda - varepsilon)$ and $C = 2$ is sufficient.

For $i > lambda + varepsilon$ it is the case that $f_i(t) leq f_i(lambda + varepsilon)$ and $C = 2$ is sufficient as well.

For $i in [lambda - varepsilon, lambda + varepsilon]$, $max_t in [lambda - varepsilon, lambda + varepsilon] f_i(t) = f_i(i) = e^-ii^i = e^i(ln(i) - 1) := M_i$. Now I can bound $e^-ii^i leq e^-(lambda - varepsilon)(lambda+varepsilon)^i$ from the definition of the range of $i$, but $e^-(lambda - varepsilon)(lambda+varepsilon)^i = e^3varepsilone^-(lambda + varepsilon)(lambda+varepsilon)^i$ so this does not control it by a constant, rather a function of $varepsilon$.

Could this come from some convexity argument ? Consider the function $h: x mapsto e^-xx^x$, and compute its second derivative $h''(x) = e^-x x^x-1 [x ln^2(x) + 1]$ which is positive on the range of interest, yielding convexity of $h$.

edited Jul 30 at 9:45

asked Jul 25 at 13:54

ippiki-ookami

303216

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Let $lambda geq 1$ and $varepsilon leq frac12sqrtlambda$.

Here is were we are at so far:

Notice that since $f_i'(t) = f_(i-1)(t)(i-t)$, $f_i$ is increasing on $[0, i]$ and decreasing on $[i, infty]$.

For $i < lambda - varepsilon$ it is the case that $f_i(t) leq f_i(lambda - varepsilon)$ and $C = 2$ is sufficient.

For $i > lambda + varepsilon$ it is the case that $f_i(t) leq f_i(lambda + varepsilon)$ and $C = 2$ is sufficient as well.

For $i in [lambda - varepsilon, lambda + varepsilon]$, $max_t in [lambda - varepsilon, lambda + varepsilon] f_i(t) = f_i(i) = e^-ii^i = e^i(ln(i) - 1) := M_i$. Now I can bound $e^-ii^i leq e^-(lambda - varepsilon)(lambda+varepsilon)^i$ from the definition of the range of $i$, but $e^-(lambda - varepsilon)(lambda+varepsilon)^i = e^3varepsilone^-(lambda + varepsilon)(lambda+varepsilon)^i$ so this does not control it by a constant, rather a function of $varepsilon$.

edited Jul 30 at 9:45

asked Jul 25 at 13:54

ippiki-ookami

303216

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up vote
3
down vote

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Let $lambda geq 1$ and $varepsilon leq frac12sqrtlambda$.

Here is were we are at so far:

Notice that since $f_i'(t) = f_(i-1)(t)(i-t)$, $f_i$ is increasing on $[0, i]$ and decreasing on $[i, infty]$.

For $i < lambda - varepsilon$ it is the case that $f_i(t) leq f_i(lambda - varepsilon)$ and $C = 2$ is sufficient.

For $i > lambda + varepsilon$ it is the case that $f_i(t) leq f_i(lambda + varepsilon)$ and $C = 2$ is sufficient as well.

For $i in [lambda - varepsilon, lambda + varepsilon]$, $max_t in [lambda - varepsilon, lambda + varepsilon] f_i(t) = f_i(i) = e^-ii^i = e^i(ln(i) - 1) := M_i$. Now I can bound $e^-ii^i leq e^-(lambda - varepsilon)(lambda+varepsilon)^i$ from the definition of the range of $i$, but $e^-(lambda - varepsilon)(lambda+varepsilon)^i = e^3varepsilone^-(lambda + varepsilon)(lambda+varepsilon)^i$ so this does not control it by a constant, rather a function of $varepsilon$.

edited Jul 30 at 9:45

asked Jul 25 at 13:54

ippiki-ookami

303216

Let $lambda geq 1$ and $varepsilon leq frac12sqrtlambda$.

Here is were we are at so far:

Notice that since $f_i'(t) = f_(i-1)(t)(i-t)$, $f_i$ is increasing on $[0, i]$ and decreasing on $[i, infty]$.

For $i < lambda - varepsilon$ it is the case that $f_i(t) leq f_i(lambda - varepsilon)$ and $C = 2$ is sufficient.

For $i > lambda + varepsilon$ it is the case that $f_i(t) leq f_i(lambda + varepsilon)$ and $C = 2$ is sufficient as well.

For $i in [lambda - varepsilon, lambda + varepsilon]$, $max_t in [lambda - varepsilon, lambda + varepsilon] f_i(t) = f_i(i) = e^-ii^i = e^i(ln(i) - 1) := M_i$. Now I can bound $e^-ii^i leq e^-(lambda - varepsilon)(lambda+varepsilon)^i$ from the definition of the range of $i$, but $e^-(lambda - varepsilon)(lambda+varepsilon)^i = e^3varepsilone^-(lambda + varepsilon)(lambda+varepsilon)^i$ so this does not control it by a constant, rather a function of $varepsilon$.

edited Jul 30 at 9:45

asked Jul 25 at 13:54

ippiki-ookami

303216

edited Jul 30 at 9:45

asked Jul 25 at 13:54

ippiki-ookami

303216

asked Jul 25 at 13:54

ippiki-ookami

303216

asked Jul 25 at 13:54

ippiki-ookami

303216

add a commentÂ |Â

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I'll change the notation up a bit, otherwise I'll get confused.

For $n=1,2,dots,$ let $f_n(x) = e^-xx^n, xge 0.$ Verify that $f_n$ is increasing on $[0,n],$ decreasing thereafter, with a global maximum value of $e^-nn^n.$

As you indicated, we only need to worry about $nin[lambda-epsilon, lambda+epsilon].$ Suppose first $nin[lambda-epsilon, lambda].$ I'll show

$$tag 1frace^-nn^ne^-(lambda-epsilon)(lambda-epsilon)^n$$

is bounded independent of $nin [lambda-epsilon,lambda],$ $lambdage 1$ and $epsilon le sqrt lambda /2.$

Now $n= lambda-epsilon +delta$ for some $delta in [0,epsilon].$ Substitute that into $(1)$ to get

$$frace^-(lambda-epsilon +delta)(lambda-epsilon +delta)^lambda-epsilon +delta e^-(lambda-epsilon)(lambda-epsilon)^lambda-epsilon +delta$$

$$tag 2= e^-delta cdot left (1+fracdeltalambda -epsilonright)^lambda -epsiloncdotleft (1+fracdeltalambda -epsilonright)^delta.$$

We need a few lemmas:

Lemma 1: If $0le ule v,$ then

$$ left(1+fracuv right )^v le e^u.$$

Lemma 2: If $uge 1,$ then

$$left(1+fracu/2u^2-u/2right)^u/2 le e^1/2.$$

I'll leave the proofs of these to you for now. Now Lemma 1 implies the second factor in $(2)$ is $le e^delta.$ As for the third factor, it only gets larger if we take $delta = epsilon=sqrt lambda/2.$ Lemma 2 then implies, with $u=sqrt lambda,$ that the third factor is no more than $e^1/2.$

So we have shown the desired boundedness in $(1),$ assuming $nin[lambda-epsilon, lambda].$ Surely the same result, using similar ideas, holds for $nin[lambda,lambda+epsilon].$ I haven't checked the latter super carefully, so it might be a good exercise for you. Assuming everything works, we have shown the desired universal $C$ exists. (And probably there is a simple estimate for $C$ that can be found.)

edited Aug 3 at 7:15

answered Jul 29 at 17:56

zhw.

65.6k42870

How does your claim imply this result ? We are interested in the existence of a universal constant $C$ such that $f_n(x) leq frac12 C left( e^-(lambda - varepsilon)(lambda - varepsilon)^n + e^-(lambda + varepsilon)(lambda + varepsilon)^n right) $ for some range of interest for $lambda$ and $epsilon$. It seems to me that your upper bound for $M_n$ introduces a $C$ that is a function of $b$ (using your notation, and $lambda + epsilon$ using mine). What did I miss ?
â€“Â ippiki-ookami
Jul 30 at 6:47

Oh. I thought $C$ was to be independent of $n$ only. Hmmm ... I'll delete this for now.
â€“Â zhw.
Jul 30 at 15:43

I've changed my answer thoroughly, and believe I've proved the desired result.
â€“Â zhw.
Aug 2 at 20:33

I agree with this proof, and award it the bounty, as it is the clearest of both and comes with the tightest constant
â€“Â ippiki-ookami
Aug 5 at 6:05

add a commentÂ |Â

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Define $f_i : [lambda - varepsilon , lambda + varepsilon ] to mathbbR, f_i(t) = e^-tt^i$. This is a continuous function with positive values. It attains a maximum $m_i > 0$. Moreover, $d_i = frac12(f_i(lambda - varepsilon) + f_i(lambda + varepsilon)) > 0$ and we find $C_i > 0$ such that $f_i(ell) le m_i < C_i d_i$ (for example $C_i = fracm_id_i +1 $).

We have $f'_i(t) = -e^-tt^i + ie^-tt^i-1 = e^-tt^i-1(i - t) = f_i-1(t)(i - t)$. Hence $f'_i(t) > 0$ for all $t$ and $i > lambda + varepsilon$.

Therefore, for $i > lambda + varepsilon$, we see that $f_i$ is strictly increasing so that $m_i = f_i(lambda + varepsilon)$ and we can take $C_i =2$.

Now take $C = max C_i $.

More detailed proof:

Define $f_n : (0,infty) to mathbbR, f_n(t) = e^-tt^n$. This is a continuous function with positive values. Moreover , define $d_n = frac12(f_n(lambda - varepsilon) + f_n(lambda + varepsilon)) > 0$.

We have $f'_n(t) = -e^-tt^n + ne^-tt^n-1 = e^-tt^n-1(n - t) = f_n-1(t)(n - t)$. Hence $f'_n(t) > 0$ for all $t < n$, $f'_n(n) = 0$ and $f'_n(t) < 0$ for all $t > n$. This means that $f_n$ is strictly increasing on $(0,n]$, attains its (global) maximum $M_n = f_n(n) = e^-n n^n = e^n(ln(n)-1)$ at $t = n$ and is strictly decreasing on $[n,infty)$.

The maximum $m_n$ of $f_n$ on the interval $[lambda - varepsilon, lambda + varepsilon]$ is therefore $m_n = f_n(lambda - varepsilon)$ for $n le lambda - varepsilon$, $m_n = M_n > 2d_n$ for $lambda - varepsilon < n < lambda + varepsilon$ and $m_n = f_n(lambda + varepsilon)$ for $n ge lambda + varepsilon$.

We conclude $f_n(ell) < f_n(lambda - varepsilon) + f_n(lambda + varepsilon) = 2d_n$ for $n notin (lambda - varepsilon, lambda + varepsilon)$.

For the finitely many $n in (lambda - varepsilon, lambda + varepsilon)$ we have $f_n(ell) le M_n = fracM_nd_n d_n$.

Define $M = max fracM_nd_n mid n in (lambda - varepsilon, lambda + varepsilon) $. We have $M > 2$. Choose any $C > M$. Then for all $n in mathbbN$

$$f_n(ell) < C d_n .$$

Added:

To show that that there exists a universal constant $C$ such that $f_n(ell) < C d_n$ for all $n in mathbbN, lambda ge 1, varepsilon in [0,frac12sqrtlambda]$ we proceed as follows.

As we know it suffices to find $C$ such that $ fracM_nd_n < C$ for all $n in (lambda - varepsilon, lambda + varepsilon)$. We claim that $C = 2 e^2$ will do.

We have
$$d_n > frac12f_n(mu)$$
where $mu = lambda - varepsilon$. Therefore
$$fracM_nd_n < 2 fracf_n(n)f_n(mu) .$$
Noting $f_n(x) = e^n ln(x) - x$ we get
$$fracf_n(n)f_n(mu) = e^n(ln(n) - ln(mu)) - (n - mu) .$$
It remains to verify that $n(ln(n) - ln(mu)) - (n - mu) < 2$, i.e.
$$n fracln(n) - ln(mu)n - mu < frac2n - mu + 1 .$$
The mean value theorem provides $x in (mu,n)$ such that
$$n fracln(n) - ln(mu)n - mu = n ln'_n(x) = n frac1x < fracnmu < fraclambda + frac12sqrtlambdalambda - frac12sqrtlambda .$$
Now $n - mu < 2varepsilon le sqrtlambda$, hence it suffices to verify
$$fraclambda + frac12sqrtlambdalambda - frac12sqrtlambda le frac2sqrtlambda + 1 .$$
This can be left as an exercise.

Note that $C = 2 e^2$ is the result of a brute force calculation. We can improve this to $C = 2 e^frac12$ as in the answer by zhw.

Repeat the above calculations for $n in (lambda - varepsilon,lambda]$ and $C = 2 e^frac12$. Then you have to verify
$$fraclambdalambda - frac12sqrtlambda le frac1sqrtlambda + 1 .$$
For $n in [lambda, lambda + varepsilon)$ replace $mu$ by $nu = lambda + varepsilon$ and verify by similar arguments
$$n(ln(n) - ln(nu)) - (n -nu) < frac12 .$$

edited Aug 3 at 15:37

answered Jul 29 at 9:58

Paul Frost

3,613420

So this takes care of $i > lambda + epsilon$. Similarly for $i < lambda - epsilon$ we can use the minimum. What about for $i in [lambda - epsilon, lambda + epsilon]$ ? (PS: small typo) in the definition of $d_i$.
â€“Â ippiki-ookami
Jul 29 at 10:54

There are only finitely many $i < lambda + varepsilon$, $i =1,...,r$ , and for these we have indivdual $C_i$. Defining $C = max C_i = max C_1,...,C_r, 2 $ we obtain $f_i(ell) le m_i le C_id_i le Cd_i$.
â€“Â Paul Frost
Jul 29 at 11:32

I agree that there are only finitely many values. But if I set $i in [lambda - epsilon; lambda + epsilon]$, then $f_i(t)$ is increasing for $t leq i$ and decreasing for $t geq i$, so the maximum will be achieved at $f_i(i)$, am I right ? Now I can compute $f_i(i)$ to be $e^i i^i = e^i (ln(i) - 1)$, which is an increasing function of $i$ since $ln(i) + 1 geq ln(1/2) + 1 > 0$, so the maximum is achieved for $i$ at $e^(lambda + epsilon) [ln(lambda + epsilon) - 1]$ (or at something of about this order since we take only integers). So is this $C_i$ really universal ?
â€“Â ippiki-ookami
Jul 29 at 12:27

1

Now I understand what you really want to prove: There exists $C >0$ such that for all $lambda, varepsilon,i, t in [lambda - varepsilon , lambda + varepsilon]$ the inequality is satisfied. This was not clear from your question where you start with given $lambda, varepsilon$.
â€“Â Paul Frost
Jul 30 at 8:51

1

Are you sure that the book claims that the stronger result is true?
â€“Â Paul Frost
Jul 30 at 8:54

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2 Answers
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I'll change the notation up a bit, otherwise I'll get confused.

For $n=1,2,dots,$ let $f_n(x) = e^-xx^n, xge 0.$ Verify that $f_n$ is increasing on $[0,n],$ decreasing thereafter, with a global maximum value of $e^-nn^n.$

As you indicated, we only need to worry about $nin[lambda-epsilon, lambda+epsilon].$ Suppose first $nin[lambda-epsilon, lambda].$ I'll show

$$tag 1frace^-nn^ne^-(lambda-epsilon)(lambda-epsilon)^n$$

is bounded independent of $nin [lambda-epsilon,lambda],$ $lambdage 1$ and $epsilon le sqrt lambda /2.$

Now $n= lambda-epsilon +delta$ for some $delta in [0,epsilon].$ Substitute that into $(1)$ to get

$$frace^-(lambda-epsilon +delta)(lambda-epsilon +delta)^lambda-epsilon +delta e^-(lambda-epsilon)(lambda-epsilon)^lambda-epsilon +delta$$

$$tag 2= e^-delta cdot left (1+fracdeltalambda -epsilonright)^lambda -epsiloncdotleft (1+fracdeltalambda -epsilonright)^delta.$$

We need a few lemmas:

Lemma 1: If $0le ule v,$ then

$$ left(1+fracuv right )^v le e^u.$$

Lemma 2: If $uge 1,$ then

$$left(1+fracu/2u^2-u/2right)^u/2 le e^1/2.$$

edited Aug 3 at 7:15

answered Jul 29 at 17:56

zhw.

65.6k42870

How does your claim imply this result ? We are interested in the existence of a universal constant $C$ such that $f_n(x) leq frac12 C left( e^-(lambda - varepsilon)(lambda - varepsilon)^n + e^-(lambda + varepsilon)(lambda + varepsilon)^n right) $ for some range of interest for $lambda$ and $epsilon$. It seems to me that your upper bound for $M_n$ introduces a $C$ that is a function of $b$ (using your notation, and $lambda + epsilon$ using mine). What did I miss ?
â€“Â ippiki-ookami
Jul 30 at 6:47

Oh. I thought $C$ was to be independent of $n$ only. Hmmm ... I'll delete this for now.
â€“Â zhw.
Jul 30 at 15:43

I've changed my answer thoroughly, and believe I've proved the desired result.
â€“Â zhw.
Aug 2 at 20:33

I agree with this proof, and award it the bounty, as it is the clearest of both and comes with the tightest constant
â€“Â ippiki-ookami
Aug 5 at 6:05

add a commentÂ |Â

up vote
2
down vote

accepted

+50

I'll change the notation up a bit, otherwise I'll get confused.

For $n=1,2,dots,$ let $f_n(x) = e^-xx^n, xge 0.$ Verify that $f_n$ is increasing on $[0,n],$ decreasing thereafter, with a global maximum value of $e^-nn^n.$

As you indicated, we only need to worry about $nin[lambda-epsilon, lambda+epsilon].$ Suppose first $nin[lambda-epsilon, lambda].$ I'll show

$$tag 1frace^-nn^ne^-(lambda-epsilon)(lambda-epsilon)^n$$

is bounded independent of $nin [lambda-epsilon,lambda],$ $lambdage 1$ and $epsilon le sqrt lambda /2.$

Now $n= lambda-epsilon +delta$ for some $delta in [0,epsilon].$ Substitute that into $(1)$ to get

$$frace^-(lambda-epsilon +delta)(lambda-epsilon +delta)^lambda-epsilon +delta e^-(lambda-epsilon)(lambda-epsilon)^lambda-epsilon +delta$$

$$tag 2= e^-delta cdot left (1+fracdeltalambda -epsilonright)^lambda -epsiloncdotleft (1+fracdeltalambda -epsilonright)^delta.$$

We need a few lemmas:

Lemma 1: If $0le ule v,$ then

$$ left(1+fracuv right )^v le e^u.$$

Lemma 2: If $uge 1,$ then

$$left(1+fracu/2u^2-u/2right)^u/2 le e^1/2.$$

edited Aug 3 at 7:15

answered Jul 29 at 17:56

zhw.

65.6k42870

How does your claim imply this result ? We are interested in the existence of a universal constant $C$ such that $f_n(x) leq frac12 C left( e^-(lambda - varepsilon)(lambda - varepsilon)^n + e^-(lambda + varepsilon)(lambda + varepsilon)^n right) $ for some range of interest for $lambda$ and $epsilon$. It seems to me that your upper bound for $M_n$ introduces a $C$ that is a function of $b$ (using your notation, and $lambda + epsilon$ using mine). What did I miss ?
â€“Â ippiki-ookami
Jul 30 at 6:47

Oh. I thought $C$ was to be independent of $n$ only. Hmmm ... I'll delete this for now.
â€“Â zhw.
Jul 30 at 15:43

I've changed my answer thoroughly, and believe I've proved the desired result.
â€“Â zhw.
Aug 2 at 20:33

I agree with this proof, and award it the bounty, as it is the clearest of both and comes with the tightest constant
â€“Â ippiki-ookami
Aug 5 at 6:05

add a commentÂ |Â

up vote
2
down vote

accepted

+50

up vote
2
down vote

accepted

+50

I'll change the notation up a bit, otherwise I'll get confused.

For $n=1,2,dots,$ let $f_n(x) = e^-xx^n, xge 0.$ Verify that $f_n$ is increasing on $[0,n],$ decreasing thereafter, with a global maximum value of $e^-nn^n.$

As you indicated, we only need to worry about $nin[lambda-epsilon, lambda+epsilon].$ Suppose first $nin[lambda-epsilon, lambda].$ I'll show

$$tag 1frace^-nn^ne^-(lambda-epsilon)(lambda-epsilon)^n$$

is bounded independent of $nin [lambda-epsilon,lambda],$ $lambdage 1$ and $epsilon le sqrt lambda /2.$

Now $n= lambda-epsilon +delta$ for some $delta in [0,epsilon].$ Substitute that into $(1)$ to get

$$frace^-(lambda-epsilon +delta)(lambda-epsilon +delta)^lambda-epsilon +delta e^-(lambda-epsilon)(lambda-epsilon)^lambda-epsilon +delta$$

$$tag 2= e^-delta cdot left (1+fracdeltalambda -epsilonright)^lambda -epsiloncdotleft (1+fracdeltalambda -epsilonright)^delta.$$

We need a few lemmas:

Lemma 1: If $0le ule v,$ then

$$ left(1+fracuv right )^v le e^u.$$

Lemma 2: If $uge 1,$ then

$$left(1+fracu/2u^2-u/2right)^u/2 le e^1/2.$$

edited Aug 3 at 7:15

answered Jul 29 at 17:56

zhw.

65.6k42870

I'll change the notation up a bit, otherwise I'll get confused.

For $n=1,2,dots,$ let $f_n(x) = e^-xx^n, xge 0.$ Verify that $f_n$ is increasing on $[0,n],$ decreasing thereafter, with a global maximum value of $e^-nn^n.$

As you indicated, we only need to worry about $nin[lambda-epsilon, lambda+epsilon].$ Suppose first $nin[lambda-epsilon, lambda].$ I'll show

$$tag 1frace^-nn^ne^-(lambda-epsilon)(lambda-epsilon)^n$$

is bounded independent of $nin [lambda-epsilon,lambda],$ $lambdage 1$ and $epsilon le sqrt lambda /2.$

Now $n= lambda-epsilon +delta$ for some $delta in [0,epsilon].$ Substitute that into $(1)$ to get

$$frace^-(lambda-epsilon +delta)(lambda-epsilon +delta)^lambda-epsilon +delta e^-(lambda-epsilon)(lambda-epsilon)^lambda-epsilon +delta$$

$$tag 2= e^-delta cdot left (1+fracdeltalambda -epsilonright)^lambda -epsiloncdotleft (1+fracdeltalambda -epsilonright)^delta.$$

We need a few lemmas:

Lemma 1: If $0le ule v,$ then

$$ left(1+fracuv right )^v le e^u.$$

Lemma 2: If $uge 1,$ then

$$left(1+fracu/2u^2-u/2right)^u/2 le e^1/2.$$

edited Aug 3 at 7:15

answered Jul 29 at 17:56

zhw.

65.6k42870

edited Aug 3 at 7:15

answered Jul 29 at 17:56

zhw.

65.6k42870

answered Jul 29 at 17:56

zhw.

65.6k42870

answered Jul 29 at 17:56

zhw.

65.6k42870

How does your claim imply this result ? We are interested in the existence of a universal constant $C$ such that $f_n(x) leq frac12 C left( e^-(lambda - varepsilon)(lambda - varepsilon)^n + e^-(lambda + varepsilon)(lambda + varepsilon)^n right) $ for some range of interest for $lambda$ and $epsilon$. It seems to me that your upper bound for $M_n$ introduces a $C$ that is a function of $b$ (using your notation, and $lambda + epsilon$ using mine). What did I miss ?
â€“Â ippiki-ookami
Jul 30 at 6:47

Oh. I thought $C$ was to be independent of $n$ only. Hmmm ... I'll delete this for now.
â€“Â zhw.
Jul 30 at 15:43

I've changed my answer thoroughly, and believe I've proved the desired result.
â€“Â zhw.
Aug 2 at 20:33

I agree with this proof, and award it the bounty, as it is the clearest of both and comes with the tightest constant
â€“Â ippiki-ookami
Aug 5 at 6:05

add a commentÂ |Â

How does your claim imply this result ? We are interested in the existence of a universal constant $C$ such that $f_n(x) leq frac12 C left( e^-(lambda - varepsilon)(lambda - varepsilon)^n + e^-(lambda + varepsilon)(lambda + varepsilon)^n right) $ for some range of interest for $lambda$ and $epsilon$. It seems to me that your upper bound for $M_n$ introduces a $C$ that is a function of $b$ (using your notation, and $lambda + epsilon$ using mine). What did I miss ?
â€“Â ippiki-ookami
Jul 30 at 6:47

Oh. I thought $C$ was to be independent of $n$ only. Hmmm ... I'll delete this for now.
â€“Â zhw.
Jul 30 at 15:43

I've changed my answer thoroughly, and believe I've proved the desired result.
â€“Â zhw.
Aug 2 at 20:33

I agree with this proof, and award it the bounty, as it is the clearest of both and comes with the tightest constant
â€“Â ippiki-ookami
Aug 5 at 6:05

How does your claim imply this result ? We are interested in the existence of a universal constant $C$ such that $f_n(x) leq frac12 C left( e^-(lambda - varepsilon)(lambda - varepsilon)^n + e^-(lambda + varepsilon)(lambda + varepsilon)^n right) $ for some range of interest for $lambda$ and $epsilon$. It seems to me that your upper bound for $M_n$ introduces a $C$ that is a function of $b$ (using your notation, and $lambda + epsilon$ using mine). What did I miss ?
â€“Â ippiki-ookami
Jul 30 at 6:47

Oh. I thought $C$ was to be independent of $n$ only. Hmmm ... I'll delete this for now.
â€“Â zhw.
Jul 30 at 15:43

I've changed my answer thoroughly, and believe I've proved the desired result.
â€“Â zhw.
Aug 2 at 20:33

I agree with this proof, and award it the bounty, as it is the clearest of both and comes with the tightest constant
â€“Â ippiki-ookami
Aug 5 at 6:05

add a commentÂ |Â

up vote
2
down vote

We have $f'_i(t) = -e^-tt^i + ie^-tt^i-1 = e^-tt^i-1(i - t) = f_i-1(t)(i - t)$. Hence $f'_i(t) > 0$ for all $t$ and $i > lambda + varepsilon$.

Therefore, for $i > lambda + varepsilon$, we see that $f_i$ is strictly increasing so that $m_i = f_i(lambda + varepsilon)$ and we can take $C_i =2$.

Now take $C = max C_i $.

More detailed proof:

Define $f_n : (0,infty) to mathbbR, f_n(t) = e^-tt^n$. This is a continuous function with positive values. Moreover , define $d_n = frac12(f_n(lambda - varepsilon) + f_n(lambda + varepsilon)) > 0$.

We conclude $f_n(ell) < f_n(lambda - varepsilon) + f_n(lambda + varepsilon) = 2d_n$ for $n notin (lambda - varepsilon, lambda + varepsilon)$.

For the finitely many $n in (lambda - varepsilon, lambda + varepsilon)$ we have $f_n(ell) le M_n = fracM_nd_n d_n$.

Define $M = max fracM_nd_n mid n in (lambda - varepsilon, lambda + varepsilon) $. We have $M > 2$. Choose any $C > M$. Then for all $n in mathbbN$

$$f_n(ell) < C d_n .$$

Added:

To show that that there exists a universal constant $C$ such that $f_n(ell) < C d_n$ for all $n in mathbbN, lambda ge 1, varepsilon in [0,frac12sqrtlambda]$ we proceed as follows.

As we know it suffices to find $C$ such that $ fracM_nd_n < C$ for all $n in (lambda - varepsilon, lambda + varepsilon)$. We claim that $C = 2 e^2$ will do.

Note that $C = 2 e^2$ is the result of a brute force calculation. We can improve this to $C = 2 e^frac12$ as in the answer by zhw.

edited Aug 3 at 15:37

answered Jul 29 at 9:58

Paul Frost

3,613420

So this takes care of $i > lambda + epsilon$. Similarly for $i < lambda - epsilon$ we can use the minimum. What about for $i in [lambda - epsilon, lambda + epsilon]$ ? (PS: small typo) in the definition of $d_i$.
â€“Â ippiki-ookami
Jul 29 at 10:54

There are only finitely many $i < lambda + varepsilon$, $i =1,...,r$ , and for these we have indivdual $C_i$. Defining $C = max C_i = max C_1,...,C_r, 2 $ we obtain $f_i(ell) le m_i le C_id_i le Cd_i$.
â€“Â Paul Frost
Jul 29 at 11:32

I agree that there are only finitely many values. But if I set $i in [lambda - epsilon; lambda + epsilon]$, then $f_i(t)$ is increasing for $t leq i$ and decreasing for $t geq i$, so the maximum will be achieved at $f_i(i)$, am I right ? Now I can compute $f_i(i)$ to be $e^i i^i = e^i (ln(i) - 1)$, which is an increasing function of $i$ since $ln(i) + 1 geq ln(1/2) + 1 > 0$, so the maximum is achieved for $i$ at $e^(lambda + epsilon) [ln(lambda + epsilon) - 1]$ (or at something of about this order since we take only integers). So is this $C_i$ really universal ?
â€“Â ippiki-ookami
Jul 29 at 12:27

1

Now I understand what you really want to prove: There exists $C >0$ such that for all $lambda, varepsilon,i, t in [lambda - varepsilon , lambda + varepsilon]$ the inequality is satisfied. This was not clear from your question where you start with given $lambda, varepsilon$.
â€“Â Paul Frost
Jul 30 at 8:51

1

Are you sure that the book claims that the stronger result is true?
â€“Â Paul Frost
Jul 30 at 8:54

Â |Â
show 4 more comments

up vote
2
down vote

We have $f'_i(t) = -e^-tt^i + ie^-tt^i-1 = e^-tt^i-1(i - t) = f_i-1(t)(i - t)$. Hence $f'_i(t) > 0$ for all $t$ and $i > lambda + varepsilon$.

Therefore, for $i > lambda + varepsilon$, we see that $f_i$ is strictly increasing so that $m_i = f_i(lambda + varepsilon)$ and we can take $C_i =2$.

Now take $C = max C_i $.

More detailed proof:

Define $f_n : (0,infty) to mathbbR, f_n(t) = e^-tt^n$. This is a continuous function with positive values. Moreover , define $d_n = frac12(f_n(lambda - varepsilon) + f_n(lambda + varepsilon)) > 0$.

We conclude $f_n(ell) < f_n(lambda - varepsilon) + f_n(lambda + varepsilon) = 2d_n$ for $n notin (lambda - varepsilon, lambda + varepsilon)$.

For the finitely many $n in (lambda - varepsilon, lambda + varepsilon)$ we have $f_n(ell) le M_n = fracM_nd_n d_n$.

Define $M = max fracM_nd_n mid n in (lambda - varepsilon, lambda + varepsilon) $. We have $M > 2$. Choose any $C > M$. Then for all $n in mathbbN$

$$f_n(ell) < C d_n .$$

Added:

To show that that there exists a universal constant $C$ such that $f_n(ell) < C d_n$ for all $n in mathbbN, lambda ge 1, varepsilon in [0,frac12sqrtlambda]$ we proceed as follows.

As we know it suffices to find $C$ such that $ fracM_nd_n < C$ for all $n in (lambda - varepsilon, lambda + varepsilon)$. We claim that $C = 2 e^2$ will do.

Note that $C = 2 e^2$ is the result of a brute force calculation. We can improve this to $C = 2 e^frac12$ as in the answer by zhw.

edited Aug 3 at 15:37

answered Jul 29 at 9:58

Paul Frost

3,613420

So this takes care of $i > lambda + epsilon$. Similarly for $i < lambda - epsilon$ we can use the minimum. What about for $i in [lambda - epsilon, lambda + epsilon]$ ? (PS: small typo) in the definition of $d_i$.
â€“Â ippiki-ookami
Jul 29 at 10:54

There are only finitely many $i < lambda + varepsilon$, $i =1,...,r$ , and for these we have indivdual $C_i$. Defining $C = max C_i = max C_1,...,C_r, 2 $ we obtain $f_i(ell) le m_i le C_id_i le Cd_i$.
â€“Â Paul Frost
Jul 29 at 11:32

I agree that there are only finitely many values. But if I set $i in [lambda - epsilon; lambda + epsilon]$, then $f_i(t)$ is increasing for $t leq i$ and decreasing for $t geq i$, so the maximum will be achieved at $f_i(i)$, am I right ? Now I can compute $f_i(i)$ to be $e^i i^i = e^i (ln(i) - 1)$, which is an increasing function of $i$ since $ln(i) + 1 geq ln(1/2) + 1 > 0$, so the maximum is achieved for $i$ at $e^(lambda + epsilon) [ln(lambda + epsilon) - 1]$ (or at something of about this order since we take only integers). So is this $C_i$ really universal ?
â€“Â ippiki-ookami
Jul 29 at 12:27

1

Now I understand what you really want to prove: There exists $C >0$ such that for all $lambda, varepsilon,i, t in [lambda - varepsilon , lambda + varepsilon]$ the inequality is satisfied. This was not clear from your question where you start with given $lambda, varepsilon$.
â€“Â Paul Frost
Jul 30 at 8:51

1

Are you sure that the book claims that the stronger result is true?
â€“Â Paul Frost
Jul 30 at 8:54

Â |Â
show 4 more comments

up vote
2
down vote

We have $f'_i(t) = -e^-tt^i + ie^-tt^i-1 = e^-tt^i-1(i - t) = f_i-1(t)(i - t)$. Hence $f'_i(t) > 0$ for all $t$ and $i > lambda + varepsilon$.

Therefore, for $i > lambda + varepsilon$, we see that $f_i$ is strictly increasing so that $m_i = f_i(lambda + varepsilon)$ and we can take $C_i =2$.

Now take $C = max C_i $.

More detailed proof:

Define $f_n : (0,infty) to mathbbR, f_n(t) = e^-tt^n$. This is a continuous function with positive values. Moreover , define $d_n = frac12(f_n(lambda - varepsilon) + f_n(lambda + varepsilon)) > 0$.

We conclude $f_n(ell) < f_n(lambda - varepsilon) + f_n(lambda + varepsilon) = 2d_n$ for $n notin (lambda - varepsilon, lambda + varepsilon)$.

For the finitely many $n in (lambda - varepsilon, lambda + varepsilon)$ we have $f_n(ell) le M_n = fracM_nd_n d_n$.

Define $M = max fracM_nd_n mid n in (lambda - varepsilon, lambda + varepsilon) $. We have $M > 2$. Choose any $C > M$. Then for all $n in mathbbN$

$$f_n(ell) < C d_n .$$

Added:

To show that that there exists a universal constant $C$ such that $f_n(ell) < C d_n$ for all $n in mathbbN, lambda ge 1, varepsilon in [0,frac12sqrtlambda]$ we proceed as follows.

As we know it suffices to find $C$ such that $ fracM_nd_n < C$ for all $n in (lambda - varepsilon, lambda + varepsilon)$. We claim that $C = 2 e^2$ will do.

Note that $C = 2 e^2$ is the result of a brute force calculation. We can improve this to $C = 2 e^frac12$ as in the answer by zhw.

edited Aug 3 at 15:37

answered Jul 29 at 9:58

Paul Frost

3,613420

We have $f'_i(t) = -e^-tt^i + ie^-tt^i-1 = e^-tt^i-1(i - t) = f_i-1(t)(i - t)$. Hence $f'_i(t) > 0$ for all $t$ and $i > lambda + varepsilon$.

Therefore, for $i > lambda + varepsilon$, we see that $f_i$ is strictly increasing so that $m_i = f_i(lambda + varepsilon)$ and we can take $C_i =2$.

Now take $C = max C_i $.

More detailed proof:

Define $f_n : (0,infty) to mathbbR, f_n(t) = e^-tt^n$. This is a continuous function with positive values. Moreover , define $d_n = frac12(f_n(lambda - varepsilon) + f_n(lambda + varepsilon)) > 0$.

We conclude $f_n(ell) < f_n(lambda - varepsilon) + f_n(lambda + varepsilon) = 2d_n$ for $n notin (lambda - varepsilon, lambda + varepsilon)$.

For the finitely many $n in (lambda - varepsilon, lambda + varepsilon)$ we have $f_n(ell) le M_n = fracM_nd_n d_n$.

Define $M = max fracM_nd_n mid n in (lambda - varepsilon, lambda + varepsilon) $. We have $M > 2$. Choose any $C > M$. Then for all $n in mathbbN$

$$f_n(ell) < C d_n .$$

Added:

To show that that there exists a universal constant $C$ such that $f_n(ell) < C d_n$ for all $n in mathbbN, lambda ge 1, varepsilon in [0,frac12sqrtlambda]$ we proceed as follows.

As we know it suffices to find $C$ such that $ fracM_nd_n < C$ for all $n in (lambda - varepsilon, lambda + varepsilon)$. We claim that $C = 2 e^2$ will do.

Note that $C = 2 e^2$ is the result of a brute force calculation. We can improve this to $C = 2 e^frac12$ as in the answer by zhw.

edited Aug 3 at 15:37

answered Jul 29 at 9:58

Paul Frost

3,613420

edited Aug 3 at 15:37

answered Jul 29 at 9:58

Paul Frost

3,613420

answered Jul 29 at 9:58

Paul Frost

3,613420

answered Jul 29 at 9:58

Paul Frost

3,613420

So this takes care of $i > lambda + epsilon$. Similarly for $i < lambda - epsilon$ we can use the minimum. What about for $i in [lambda - epsilon, lambda + epsilon]$ ? (PS: small typo) in the definition of $d_i$.
â€“Â ippiki-ookami
Jul 29 at 10:54

There are only finitely many $i < lambda + varepsilon$, $i =1,...,r$ , and for these we have indivdual $C_i$. Defining $C = max C_i = max C_1,...,C_r, 2 $ we obtain $f_i(ell) le m_i le C_id_i le Cd_i$.
â€“Â Paul Frost
Jul 29 at 11:32

I agree that there are only finitely many values. But if I set $i in [lambda - epsilon; lambda + epsilon]$, then $f_i(t)$ is increasing for $t leq i$ and decreasing for $t geq i$, so the maximum will be achieved at $f_i(i)$, am I right ? Now I can compute $f_i(i)$ to be $e^i i^i = e^i (ln(i) - 1)$, which is an increasing function of $i$ since $ln(i) + 1 geq ln(1/2) + 1 > 0$, so the maximum is achieved for $i$ at $e^(lambda + epsilon) [ln(lambda + epsilon) - 1]$ (or at something of about this order since we take only integers). So is this $C_i$ really universal ?
â€“Â ippiki-ookami
Jul 29 at 12:27

1

Now I understand what you really want to prove: There exists $C >0$ such that for all $lambda, varepsilon,i, t in [lambda - varepsilon , lambda + varepsilon]$ the inequality is satisfied. This was not clear from your question where you start with given $lambda, varepsilon$.
â€“Â Paul Frost
Jul 30 at 8:51

1

Are you sure that the book claims that the stronger result is true?
â€“Â Paul Frost
Jul 30 at 8:54

Â |Â
show 4 more comments

So this takes care of $i > lambda + epsilon$. Similarly for $i < lambda - epsilon$ we can use the minimum. What about for $i in [lambda - epsilon, lambda + epsilon]$ ? (PS: small typo) in the definition of $d_i$.
â€“Â ippiki-ookami
Jul 29 at 10:54

There are only finitely many $i < lambda + varepsilon$, $i =1,...,r$ , and for these we have indivdual $C_i$. Defining $C = max C_i = max C_1,...,C_r, 2 $ we obtain $f_i(ell) le m_i le C_id_i le Cd_i$.
â€“Â Paul Frost
Jul 29 at 11:32

I agree that there are only finitely many values. But if I set $i in [lambda - epsilon; lambda + epsilon]$, then $f_i(t)$ is increasing for $t leq i$ and decreasing for $t geq i$, so the maximum will be achieved at $f_i(i)$, am I right ? Now I can compute $f_i(i)$ to be $e^i i^i = e^i (ln(i) - 1)$, which is an increasing function of $i$ since $ln(i) + 1 geq ln(1/2) + 1 > 0$, so the maximum is achieved for $i$ at $e^(lambda + epsilon) [ln(lambda + epsilon) - 1]$ (or at something of about this order since we take only integers). So is this $C_i$ really universal ?
â€“Â ippiki-ookami
Jul 29 at 12:27

1

Now I understand what you really want to prove: There exists $C >0$ such that for all $lambda, varepsilon,i, t in [lambda - varepsilon , lambda + varepsilon]$ the inequality is satisfied. This was not clear from your question where you start with given $lambda, varepsilon$.
â€“Â Paul Frost
Jul 30 at 8:51

1

Are you sure that the book claims that the stronger result is true?
â€“Â Paul Frost
Jul 30 at 8:54

So this takes care of $i > lambda + epsilon$. Similarly for $i < lambda - epsilon$ we can use the minimum. What about for $i in [lambda - epsilon, lambda + epsilon]$ ? (PS: small typo) in the definition of $d_i$.
â€“Â ippiki-ookami
Jul 29 at 10:54

There are only finitely many $i < lambda + varepsilon$, $i =1,...,r$ , and for these we have indivdual $C_i$. Defining $C = max C_i = max C_1,...,C_r, 2 $ we obtain $f_i(ell) le m_i le C_id_i le Cd_i$.
â€“Â Paul Frost
Jul 29 at 11:32

I agree that there are only finitely many values. But if I set $i in [lambda - epsilon; lambda + epsilon]$, then $f_i(t)$ is increasing for $t leq i$ and decreasing for $t geq i$, so the maximum will be achieved at $f_i(i)$, am I right ? Now I can compute $f_i(i)$ to be $e^i i^i = e^i (ln(i) - 1)$, which is an increasing function of $i$ since $ln(i) + 1 geq ln(1/2) + 1 > 0$, so the maximum is achieved for $i$ at $e^(lambda + epsilon) [ln(lambda + epsilon) - 1]$ (or at something of about this order since we take only integers). So is this $C_i$ really universal ?
â€“Â ippiki-ookami
Jul 29 at 12:27

Now I understand what you really want to prove: There exists $C >0$ such that for all $lambda, varepsilon,i, t in [lambda - varepsilon , lambda + varepsilon]$ the inequality is satisfied. This was not clear from your question where you start with given $lambda, varepsilon$.
â€“Â Paul Frost
Jul 30 at 8:51

Are you sure that the book claims that the stronger result is true?
â€“Â Paul Frost
Jul 30 at 8:54

Â |Â
show 4 more comments

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