Mixing
Upper bound for an expression involving an integral
Clash Royale CLAN TAG#URR8PPP
up vote
1
down vote
favorite
Let $f$ be a function such that $|f(u)-f(v)| leq |u-v|$ for all $u$ and $v$ in the interval $[a,b]$.
Prove that
$$left|int_a^b f(x) dx- (b-a)f(c)right| leqfrac(b-a)^22$$
where $cin [a,b]$.
Is f continuous?
Any hint or suggestion will be appreciated.
calculus real-analysis
asked Jul 15 at 11:53
blue boy
569211
add a comment |Â
up vote
1
down vote
favorite
Let $f$ be a function such that $|f(u)-f(v)| leq |u-v|$ for all $u$ and $v$ in the interval $[a,b]$.
Prove that
$$left|int_a^b f(x) dx- (b-a)f(c)right| leqfrac(b-a)^22$$
where $cin [a,b]$.
Is f continuous?
Any hint or suggestion will be appreciated.
calculus real-analysis
asked Jul 15 at 11:53
blue boy
569211
What is $c$? What did you try?
– Robert Z
Jul 15 at 11:54
I suppose $cin [a,b]$, no ? By the way, what is the connection between the question and the title ?
– Surb
Jul 15 at 11:55
1
What has the question to do with the title?
– José Carlos Santos
Jul 15 at 12:00
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Let $f$ be a function such that $|f(u)-f(v)| leq |u-v|$ for all $u$ and $v$ in the interval $[a,b]$.
Prove that
$$left|int_a^b f(x) dx- (b-a)f(c)right| leqfrac(b-a)^22$$
where $cin [a,b]$.
Is f continuous?
Any hint or suggestion will be appreciated.
calculus real-analysis
asked Jul 15 at 11:53
blue boy
569211
Let $f$ be a function such that $|f(u)-f(v)| leq |u-v|$ for all $u$ and $v$ in the interval $[a,b]$.
Prove that
$$left|int_a^b f(x) dx- (b-a)f(c)right| leqfrac(b-a)^22$$
where $cin [a,b]$.
Is f continuous?
Any hint or suggestion will be appreciated.
calculus real-analysis
asked Jul 15 at 11:53
blue boy
569211
edited Jul 16 at 1:59
asked Jul 15 at 11:53
blue boy
569211
asked Jul 15 at 11:53
blue boy
569211
asked Jul 15 at 11:53
blue boy
569211
569211
What is $c$? What did you try?
– Robert Z
Jul 15 at 11:54
I suppose $cin [a,b]$, no ? By the way, what is the connection between the question and the title ?
– Surb
Jul 15 at 11:55
1
What has the question to do with the title?
– José Carlos Santos
Jul 15 at 12:00
add a comment |Â
What is $c$? What did you try?
– Robert Z
Jul 15 at 11:54
I suppose $cin [a,b]$, no ? By the way, what is the connection between the question and the title ?
– Surb
Jul 15 at 11:55
1
What has the question to do with the title?
– José Carlos Santos
Jul 15 at 12:00
What is $c$? What did you try?
– Robert Z
Jul 15 at 11:54
What is $c$? What did you try?
– Robert Z
Jul 15 at 11:54
I suppose $cin [a,b]$, no ? By the way, what is the connection between the question and the title ?
– Surb
Jul 15 at 11:55
I suppose $cin [a,b]$, no ? By the way, what is the connection between the question and the title ?
– Surb
Jul 15 at 11:55
1
1
What has the question to do with the title?
– José Carlos Santos
Jul 15 at 12:00
What has the question to do with the title?
– José Carlos Santos
Jul 15 at 12:00
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
4
down vote
Hint
$$f(c)(b-a)=int_a^b f(c)dx.$$
answered Jul 15 at 11:54
Surb
36.3k84274
add a comment |Â
up vote
2
down vote
As @Surd suggests, we have
beginalign
left|int_a^b f(x),dx - f(c)(b-a)right| &= left|int_a^b f(x),dx - int_a^b f(c),dxright|\
&= left|int_a^b (f(x) - f(c)),dxright|\
&le int_a^b |f(x) - f(c)|,dx\
&le int_a^b |x-c|,dx\
&= int_a^c(c-x),dx + int_c^b (x-c),dx\
&= frac12(c-a)^2 + frac12(b-c)^2\
&le frac12(b-a)^2
endalign
because
$$(c-a)^2 + (b-c)^2 le (c-a)^2 + (b-c)^2 + underbrace2(c-a)(b-c)_ge0 = big((c-a) + (b-c)big)^2 = (b-a)^2$$
answered Jul 15 at 13:05
mechanodroid
22.3k52041
You could also differentiate $ctofrac12(c-a)^2 + frac12(b-c)^2$ to find the minimum is when $c$ is the midpoint (which is pleasing intuitively). Thus the max occurs at an endpoint.
– zhw.
Jul 15 at 14:57
@zhw Yes, the minimum value also follows from AM-QM: $$sqrtfrac(c-a)^2+(b-c)^22 ge frac(c-a) + (b-c)2 = fracb-a2$$ with equality iff $c = fraca+b2$. I'm not sure how this implies anything about maxima.
– mechanodroid
Jul 15 at 18:32
@pabodu No, the minimum of the function $cmapstofrac12(c-a)^2 + frac12(b-c)^2$ is $frac(b-a)^24$ and it is attained for $c = fraca+b2$. The maximum is $frac(b-a)^22$ and it is attained for $c = a, b$.
– mechanodroid
Jul 15 at 19:08
Is f continuous
– blue boy
Jul 16 at 1:56
@blueboy Yes it is. For $varepsilon > 0$ set $delta = varepsilon$. We have $$|x-y| < delta implies |f(x) - f(y)| le |x-y| < delta = varepsilon$$ for all $x,y in [a,b]$.
– mechanodroid
Jul 16 at 8:51
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
4
down vote
Hint
$$f(c)(b-a)=int_a^b f(c)dx.$$
answered Jul 15 at 11:54
Surb
36.3k84274
add a comment |Â
up vote
4
down vote
Hint
$$f(c)(b-a)=int_a^b f(c)dx.$$
answered Jul 15 at 11:54
Surb
36.3k84274
add a comment |Â
up vote
4
down vote
up vote
4
down vote
Hint
$$f(c)(b-a)=int_a^b f(c)dx.$$
answered Jul 15 at 11:54
Surb
36.3k84274
Hint
$$f(c)(b-a)=int_a^b f(c)dx.$$
answered Jul 15 at 11:54
Surb
36.3k84274
answered Jul 15 at 11:54
Surb
36.3k84274
answered Jul 15 at 11:54
Surb
36.3k84274
answered Jul 15 at 11:54
Surb
36.3k84274
36.3k84274
add a comment |Â
add a comment |Â
up vote
2
down vote
As @Surd suggests, we have
beginalign
left|int_a^b f(x),dx - f(c)(b-a)right| &= left|int_a^b f(x),dx - int_a^b f(c),dxright|\
&= left|int_a^b (f(x) - f(c)),dxright|\
&le int_a^b |f(x) - f(c)|,dx\
&le int_a^b |x-c|,dx\
&= int_a^c(c-x),dx + int_c^b (x-c),dx\
&= frac12(c-a)^2 + frac12(b-c)^2\
&le frac12(b-a)^2
endalign
because
$$(c-a)^2 + (b-c)^2 le (c-a)^2 + (b-c)^2 + underbrace2(c-a)(b-c)_ge0 = big((c-a) + (b-c)big)^2 = (b-a)^2$$
answered Jul 15 at 13:05
mechanodroid
22.3k52041
You could also differentiate $ctofrac12(c-a)^2 + frac12(b-c)^2$ to find the minimum is when $c$ is the midpoint (which is pleasing intuitively). Thus the max occurs at an endpoint.
– zhw.
Jul 15 at 14:57
@zhw Yes, the minimum value also follows from AM-QM: $$sqrtfrac(c-a)^2+(b-c)^22 ge frac(c-a) + (b-c)2 = fracb-a2$$ with equality iff $c = fraca+b2$. I'm not sure how this implies anything about maxima.
– mechanodroid
Jul 15 at 18:32
@pabodu No, the minimum of the function $cmapstofrac12(c-a)^2 + frac12(b-c)^2$ is $frac(b-a)^24$ and it is attained for $c = fraca+b2$. The maximum is $frac(b-a)^22$ and it is attained for $c = a, b$.
– mechanodroid
Jul 15 at 19:08
Is f continuous
– blue boy
Jul 16 at 1:56
@blueboy Yes it is. For $varepsilon > 0$ set $delta = varepsilon$. We have $$|x-y| < delta implies |f(x) - f(y)| le |x-y| < delta = varepsilon$$ for all $x,y in [a,b]$.
– mechanodroid
Jul 16 at 8:51
add a comment |Â
up vote
2
down vote
As @Surd suggests, we have
beginalign
left|int_a^b f(x),dx - f(c)(b-a)right| &= left|int_a^b f(x),dx - int_a^b f(c),dxright|\
&= left|int_a^b (f(x) - f(c)),dxright|\
&le int_a^b |f(x) - f(c)|,dx\
&le int_a^b |x-c|,dx\
&= int_a^c(c-x),dx + int_c^b (x-c),dx\
&= frac12(c-a)^2 + frac12(b-c)^2\
&le frac12(b-a)^2
endalign
because
$$(c-a)^2 + (b-c)^2 le (c-a)^2 + (b-c)^2 + underbrace2(c-a)(b-c)_ge0 = big((c-a) + (b-c)big)^2 = (b-a)^2$$
answered Jul 15 at 13:05
mechanodroid
22.3k52041
You could also differentiate $ctofrac12(c-a)^2 + frac12(b-c)^2$ to find the minimum is when $c$ is the midpoint (which is pleasing intuitively). Thus the max occurs at an endpoint.
– zhw.
Jul 15 at 14:57
@zhw Yes, the minimum value also follows from AM-QM: $$sqrtfrac(c-a)^2+(b-c)^22 ge frac(c-a) + (b-c)2 = fracb-a2$$ with equality iff $c = fraca+b2$. I'm not sure how this implies anything about maxima.
– mechanodroid
Jul 15 at 18:32
@pabodu No, the minimum of the function $cmapstofrac12(c-a)^2 + frac12(b-c)^2$ is $frac(b-a)^24$ and it is attained for $c = fraca+b2$. The maximum is $frac(b-a)^22$ and it is attained for $c = a, b$.
– mechanodroid
Jul 15 at 19:08
Is f continuous
– blue boy
Jul 16 at 1:56
@blueboy Yes it is. For $varepsilon > 0$ set $delta = varepsilon$. We have $$|x-y| < delta implies |f(x) - f(y)| le |x-y| < delta = varepsilon$$ for all $x,y in [a,b]$.
– mechanodroid
Jul 16 at 8:51
add a comment |Â
up vote
2
down vote
up vote
2
down vote
As @Surd suggests, we have
beginalign
left|int_a^b f(x),dx - f(c)(b-a)right| &= left|int_a^b f(x),dx - int_a^b f(c),dxright|\
&= left|int_a^b (f(x) - f(c)),dxright|\
&le int_a^b |f(x) - f(c)|,dx\
&le int_a^b |x-c|,dx\
&= int_a^c(c-x),dx + int_c^b (x-c),dx\
&= frac12(c-a)^2 + frac12(b-c)^2\
&le frac12(b-a)^2
endalign
because
$$(c-a)^2 + (b-c)^2 le (c-a)^2 + (b-c)^2 + underbrace2(c-a)(b-c)_ge0 = big((c-a) + (b-c)big)^2 = (b-a)^2$$
answered Jul 15 at 13:05
mechanodroid
22.3k52041
As @Surd suggests, we have
beginalign
left|int_a^b f(x),dx - f(c)(b-a)right| &= left|int_a^b f(x),dx - int_a^b f(c),dxright|\
&= left|int_a^b (f(x) - f(c)),dxright|\
&le int_a^b |f(x) - f(c)|,dx\
&le int_a^b |x-c|,dx\
&= int_a^c(c-x),dx + int_c^b (x-c),dx\
&= frac12(c-a)^2 + frac12(b-c)^2\
&le frac12(b-a)^2
endalign
because
$$(c-a)^2 + (b-c)^2 le (c-a)^2 + (b-c)^2 + underbrace2(c-a)(b-c)_ge0 = big((c-a) + (b-c)big)^2 = (b-a)^2$$
answered Jul 15 at 13:05
mechanodroid
22.3k52041
answered Jul 15 at 13:05
mechanodroid
22.3k52041
answered Jul 15 at 13:05
mechanodroid
22.3k52041
answered Jul 15 at 13:05
mechanodroid
22.3k52041
22.3k52041
You could also differentiate $ctofrac12(c-a)^2 + frac12(b-c)^2$ to find the minimum is when $c$ is the midpoint (which is pleasing intuitively). Thus the max occurs at an endpoint.
– zhw.
Jul 15 at 14:57
@zhw Yes, the minimum value also follows from AM-QM: $$sqrtfrac(c-a)^2+(b-c)^22 ge frac(c-a) + (b-c)2 = fracb-a2$$ with equality iff $c = fraca+b2$. I'm not sure how this implies anything about maxima.
– mechanodroid
Jul 15 at 18:32
@pabodu No, the minimum of the function $cmapstofrac12(c-a)^2 + frac12(b-c)^2$ is $frac(b-a)^24$ and it is attained for $c = fraca+b2$. The maximum is $frac(b-a)^22$ and it is attained for $c = a, b$.
– mechanodroid
Jul 15 at 19:08
Is f continuous
– blue boy
Jul 16 at 1:56
@blueboy Yes it is. For $varepsilon > 0$ set $delta = varepsilon$. We have $$|x-y| < delta implies |f(x) - f(y)| le |x-y| < delta = varepsilon$$ for all $x,y in [a,b]$.
– mechanodroid
Jul 16 at 8:51
add a comment |Â
You could also differentiate $ctofrac12(c-a)^2 + frac12(b-c)^2$ to find the minimum is when $c$ is the midpoint (which is pleasing intuitively). Thus the max occurs at an endpoint.
– zhw.
Jul 15 at 14:57
@zhw Yes, the minimum value also follows from AM-QM: $$sqrtfrac(c-a)^2+(b-c)^22 ge frac(c-a) + (b-c)2 = fracb-a2$$ with equality iff $c = fraca+b2$. I'm not sure how this implies anything about maxima.
– mechanodroid
Jul 15 at 18:32
@pabodu No, the minimum of the function $cmapstofrac12(c-a)^2 + frac12(b-c)^2$ is $frac(b-a)^24$ and it is attained for $c = fraca+b2$. The maximum is $frac(b-a)^22$ and it is attained for $c = a, b$.
– mechanodroid
Jul 15 at 19:08
Is f continuous
– blue boy
Jul 16 at 1:56
@blueboy Yes it is. For $varepsilon > 0$ set $delta = varepsilon$. We have $$|x-y| < delta implies |f(x) - f(y)| le |x-y| < delta = varepsilon$$ for all $x,y in [a,b]$.
– mechanodroid
Jul 16 at 8:51
You could also differentiate $ctofrac12(c-a)^2 + frac12(b-c)^2$ to find the minimum is when $c$ is the midpoint (which is pleasing intuitively). Thus the max occurs at an endpoint.
– zhw.
Jul 15 at 14:57
You could also differentiate $ctofrac12(c-a)^2 + frac12(b-c)^2$ to find the minimum is when $c$ is the midpoint (which is pleasing intuitively). Thus the max occurs at an endpoint.
– zhw.
Jul 15 at 14:57
@zhw Yes, the minimum value also follows from AM-QM: $$sqrtfrac(c-a)^2+(b-c)^22 ge frac(c-a) + (b-c)2 = fracb-a2$$ with equality iff $c = fraca+b2$. I'm not sure how this implies anything about maxima.
– mechanodroid
Jul 15 at 18:32
@zhw Yes, the minimum value also follows from AM-QM: $$sqrtfrac(c-a)^2+(b-c)^22 ge frac(c-a) + (b-c)2 = fracb-a2$$ with equality iff $c = fraca+b2$. I'm not sure how this implies anything about maxima.
– mechanodroid
Jul 15 at 18:32
@pabodu No, the minimum of the function $cmapstofrac12(c-a)^2 + frac12(b-c)^2$ is $frac(b-a)^24$ and it is attained for $c = fraca+b2$. The maximum is $frac(b-a)^22$ and it is attained for $c = a, b$.
– mechanodroid
Jul 15 at 19:08
@pabodu No, the minimum of the function $cmapstofrac12(c-a)^2 + frac12(b-c)^2$ is $frac(b-a)^24$ and it is attained for $c = fraca+b2$. The maximum is $frac(b-a)^22$ and it is attained for $c = a, b$.
– mechanodroid
Jul 15 at 19:08
Is f continuous
– blue boy
Jul 16 at 1:56
Is f continuous
– blue boy
Jul 16 at 1:56
@blueboy Yes it is. For $varepsilon > 0$ set $delta = varepsilon$. We have $$|x-y| < delta implies |f(x) - f(y)| le |x-y| < delta = varepsilon$$ for all $x,y in [a,b]$.
– mechanodroid
Jul 16 at 8:51
@blueboy Yes it is. For $varepsilon > 0$ set $delta = varepsilon$. We have $$|x-y| < delta implies |f(x) - f(y)| le |x-y| < delta = varepsilon$$ for all $x,y in [a,b]$.
– mechanodroid
Jul 16 at 8:51
add a comment |Â
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2852446%2fupper-bound-for-an-expression-involving-an-integral%23new-answer', 'question_page');
);
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
What is $c$? What did you try?
– Robert Z
Jul 15 at 11:54
I suppose $cin [a,b]$, no ? By the way, what is the connection between the question and the title ?
– Surb
Jul 15 at 11:55
1
What has the question to do with the title?
– José Carlos Santos
Jul 15 at 12:00