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Linear function on space of smooth functions, Taylor series
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Let $Q:C^infty(mathbbR^n)tomathbbR$ be a linear function with the property that if $f(0)=0$ and $fgeq 0$ in a neighborhood of $0$, then $Q(f)geq 0$. Show that there are constants $c, a_i, b)ij$, where $1leq i,jleq n$ such that $Q(f)=cf(0)+sum_ifracpartial fpartial x_i(0)a_i+sum_i,jfracpartial^2fpartial x_ipartial x_j(0)b_ij$
My attempt: for each $xinmathbbR^n$, expand $f$ up to third order $f(x)=f(0)+sum_ifracpartial fpartial x_i(0)x_i+sum_i,jfracpartial^2 fpartial x_ipartial x_j(0)x_ix_j+sum_i,j,kfracpartial^3 fpartial x_ipartial x_jpartial x_k(zeta)x_ix_jx_k$, where $zeta$ is some point on the line segment connecting $0$ and $x$. Let $pi:mathbbR^ntomathbbR$ denote the projection onto the ith coordinate. $pi$ is smooth. Let $Q(pi_i)=a_i$ and $Q(pi_ipi_j)=b_ij$. We can write $f=f(0)+sum_ifracpartial fpartial x_i(0)pi_i+sum_i,jfracpartial^2 fpartial x_ipartial x_j(0)pi_ipi_j+sum_i,j,kfracpartial^3 fpartial x_ipartial x_jpartial x_k(zeta)pi_ipi_jpi_k$. Using the linearity of of $Q$, we have $Q(f)=cf(0)+sum_ifracpartial fpartial x_i(0)a_i+sum_i,jfracpartial^2 fpartial x_ipartial x_j(0)b_ij+sum_i,j,kfracpartial^3 fpartial x_ipartial x_jpartial x_k(zeta)Q(pi_ipi_jpi_k)$. Now I'd like to show that $Q(pi_ipi_jpi_k)=0$, but I'm not really seeing how. $pi_ipi_jpi_k(0)=0$, but this function is not positive in any neighborhood of $0$. Could someone please offer some insight? Thank you
calculus real-analysis
asked Jul 23 at 11:20
Tom Chalmer
261212
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1
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Let $Q:C^infty(mathbbR^n)tomathbbR$ be a linear function with the property that if $f(0)=0$ and $fgeq 0$ in a neighborhood of $0$, then $Q(f)geq 0$. Show that there are constants $c, a_i, b)ij$, where $1leq i,jleq n$ such that $Q(f)=cf(0)+sum_ifracpartial fpartial x_i(0)a_i+sum_i,jfracpartial^2fpartial x_ipartial x_j(0)b_ij$
My attempt: for each $xinmathbbR^n$, expand $f$ up to third order $f(x)=f(0)+sum_ifracpartial fpartial x_i(0)x_i+sum_i,jfracpartial^2 fpartial x_ipartial x_j(0)x_ix_j+sum_i,j,kfracpartial^3 fpartial x_ipartial x_jpartial x_k(zeta)x_ix_jx_k$, where $zeta$ is some point on the line segment connecting $0$ and $x$. Let $pi:mathbbR^ntomathbbR$ denote the projection onto the ith coordinate. $pi$ is smooth. Let $Q(pi_i)=a_i$ and $Q(pi_ipi_j)=b_ij$. We can write $f=f(0)+sum_ifracpartial fpartial x_i(0)pi_i+sum_i,jfracpartial^2 fpartial x_ipartial x_j(0)pi_ipi_j+sum_i,j,kfracpartial^3 fpartial x_ipartial x_jpartial x_k(zeta)pi_ipi_jpi_k$. Using the linearity of of $Q$, we have $Q(f)=cf(0)+sum_ifracpartial fpartial x_i(0)a_i+sum_i,jfracpartial^2 fpartial x_ipartial x_j(0)b_ij+sum_i,j,kfracpartial^3 fpartial x_ipartial x_jpartial x_k(zeta)Q(pi_ipi_jpi_k)$. Now I'd like to show that $Q(pi_ipi_jpi_k)=0$, but I'm not really seeing how. $pi_ipi_jpi_k(0)=0$, but this function is not positive in any neighborhood of $0$. Could someone please offer some insight? Thank you
calculus real-analysis
asked Jul 23 at 11:20
Tom Chalmer
261212
1
The "Using the linearity of $Q$..." is not right, because $zeta$ is not a scalar! In fact $zeta$ is a function of $x$...
– David C. Ullrich
Jul 23 at 15:58
Ah yes, good point
– Tom Chalmer
Jul 23 at 17:28
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Let $Q:C^infty(mathbbR^n)tomathbbR$ be a linear function with the property that if $f(0)=0$ and $fgeq 0$ in a neighborhood of $0$, then $Q(f)geq 0$. Show that there are constants $c, a_i, b)ij$, where $1leq i,jleq n$ such that $Q(f)=cf(0)+sum_ifracpartial fpartial x_i(0)a_i+sum_i,jfracpartial^2fpartial x_ipartial x_j(0)b_ij$
My attempt: for each $xinmathbbR^n$, expand $f$ up to third order $f(x)=f(0)+sum_ifracpartial fpartial x_i(0)x_i+sum_i,jfracpartial^2 fpartial x_ipartial x_j(0)x_ix_j+sum_i,j,kfracpartial^3 fpartial x_ipartial x_jpartial x_k(zeta)x_ix_jx_k$, where $zeta$ is some point on the line segment connecting $0$ and $x$. Let $pi:mathbbR^ntomathbbR$ denote the projection onto the ith coordinate. $pi$ is smooth. Let $Q(pi_i)=a_i$ and $Q(pi_ipi_j)=b_ij$. We can write $f=f(0)+sum_ifracpartial fpartial x_i(0)pi_i+sum_i,jfracpartial^2 fpartial x_ipartial x_j(0)pi_ipi_j+sum_i,j,kfracpartial^3 fpartial x_ipartial x_jpartial x_k(zeta)pi_ipi_jpi_k$. Using the linearity of of $Q$, we have $Q(f)=cf(0)+sum_ifracpartial fpartial x_i(0)a_i+sum_i,jfracpartial^2 fpartial x_ipartial x_j(0)b_ij+sum_i,j,kfracpartial^3 fpartial x_ipartial x_jpartial x_k(zeta)Q(pi_ipi_jpi_k)$. Now I'd like to show that $Q(pi_ipi_jpi_k)=0$, but I'm not really seeing how. $pi_ipi_jpi_k(0)=0$, but this function is not positive in any neighborhood of $0$. Could someone please offer some insight? Thank you
calculus real-analysis
asked Jul 23 at 11:20
Tom Chalmer
261212
Let $Q:C^infty(mathbbR^n)tomathbbR$ be a linear function with the property that if $f(0)=0$ and $fgeq 0$ in a neighborhood of $0$, then $Q(f)geq 0$. Show that there are constants $c, a_i, b)ij$, where $1leq i,jleq n$ such that $Q(f)=cf(0)+sum_ifracpartial fpartial x_i(0)a_i+sum_i,jfracpartial^2fpartial x_ipartial x_j(0)b_ij$
My attempt: for each $xinmathbbR^n$, expand $f$ up to third order $f(x)=f(0)+sum_ifracpartial fpartial x_i(0)x_i+sum_i,jfracpartial^2 fpartial x_ipartial x_j(0)x_ix_j+sum_i,j,kfracpartial^3 fpartial x_ipartial x_jpartial x_k(zeta)x_ix_jx_k$, where $zeta$ is some point on the line segment connecting $0$ and $x$. Let $pi:mathbbR^ntomathbbR$ denote the projection onto the ith coordinate. $pi$ is smooth. Let $Q(pi_i)=a_i$ and $Q(pi_ipi_j)=b_ij$. We can write $f=f(0)+sum_ifracpartial fpartial x_i(0)pi_i+sum_i,jfracpartial^2 fpartial x_ipartial x_j(0)pi_ipi_j+sum_i,j,kfracpartial^3 fpartial x_ipartial x_jpartial x_k(zeta)pi_ipi_jpi_k$. Using the linearity of of $Q$, we have $Q(f)=cf(0)+sum_ifracpartial fpartial x_i(0)a_i+sum_i,jfracpartial^2 fpartial x_ipartial x_j(0)b_ij+sum_i,j,kfracpartial^3 fpartial x_ipartial x_jpartial x_k(zeta)Q(pi_ipi_jpi_k)$. Now I'd like to show that $Q(pi_ipi_jpi_k)=0$, but I'm not really seeing how. $pi_ipi_jpi_k(0)=0$, but this function is not positive in any neighborhood of $0$. Could someone please offer some insight? Thank you
calculus real-analysis
asked Jul 23 at 11:20
Tom Chalmer
261212
asked Jul 23 at 11:20
Tom Chalmer
261212
asked Jul 23 at 11:20
Tom Chalmer
261212
asked Jul 23 at 11:20
Tom Chalmer
261212
261212
1
The "Using the linearity of $Q$..." is not right, because $zeta$ is not a scalar! In fact $zeta$ is a function of $x$...
– David C. Ullrich
Jul 23 at 15:58
Ah yes, good point
– Tom Chalmer
Jul 23 at 17:28
add a comment |Â
1
The "Using the linearity of $Q$..." is not right, because $zeta$ is not a scalar! In fact $zeta$ is a function of $x$...
– David C. Ullrich
Jul 23 at 15:58
Ah yes, good point
– Tom Chalmer
Jul 23 at 17:28
1
1
The "Using the linearity of $Q$..." is not right, because $zeta$ is not a scalar! In fact $zeta$ is a function of $x$...
– David C. Ullrich
Jul 23 at 15:58
The "Using the linearity of $Q$..." is not right, because $zeta$ is not a scalar! In fact $zeta$ is a function of $x$...
– David C. Ullrich
Jul 23 at 15:58
Ah yes, good point
– Tom Chalmer
Jul 23 at 17:28
Ah yes, good point
– Tom Chalmer
Jul 23 at 17:28
add a comment |Â
1 Answer
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That's curious. I'm going to consider the case $n=1$.
Lemma. If $f(0)=f'(0)=f''(0)=0$ then $Q f=0$.
Proof. For $epsilon>0$ let $$f_epsilon(x)=f(x)+epsilon x^2.$$
Taylor's theorem shows that $f(x)=o(x^2)$ as $xto0$, so for every $epsilon>0$ we have $f_epsilonge0$ in a neighborhood of the origin. Since $f_epsilon(0)=0$ this shows that $$Qf_epsilonge0quad(epsilon>0).$$
So if $g(x)=x^2$ and $alpha=Qg$ we have $$Qf+epsilonalphage0$$for every $epsilon>0$, hence $$Qfge0.$$
But $-f$ satisfies the same hypothesis, so we also have $$-Qfge0,$$hence $Qf=0$.
Outline of how the result follows: Say $P$ is the space of polynomials of degree no larger than $2$. Show that there exist $a,b,c$ so that $$Qp=ap(0)+bp'(0)+cp''(0)quad(pin P).$$
Now for every $f$ there exists $pin P$ with $p(0)=f(0)$, $p'(0)=f'(0)$ and $p''(0)=f''(0)$. The lemma shows that $Qf=Qp$.
answered Jul 23 at 15:53
David C. Ullrich
54.1k33481
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
That's curious. I'm going to consider the case $n=1$.
Lemma. If $f(0)=f'(0)=f''(0)=0$ then $Q f=0$.
Proof. For $epsilon>0$ let $$f_epsilon(x)=f(x)+epsilon x^2.$$
Taylor's theorem shows that $f(x)=o(x^2)$ as $xto0$, so for every $epsilon>0$ we have $f_epsilonge0$ in a neighborhood of the origin. Since $f_epsilon(0)=0$ this shows that $$Qf_epsilonge0quad(epsilon>0).$$
So if $g(x)=x^2$ and $alpha=Qg$ we have $$Qf+epsilonalphage0$$for every $epsilon>0$, hence $$Qfge0.$$
But $-f$ satisfies the same hypothesis, so we also have $$-Qfge0,$$hence $Qf=0$.
Outline of how the result follows: Say $P$ is the space of polynomials of degree no larger than $2$. Show that there exist $a,b,c$ so that $$Qp=ap(0)+bp'(0)+cp''(0)quad(pin P).$$
Now for every $f$ there exists $pin P$ with $p(0)=f(0)$, $p'(0)=f'(0)$ and $p''(0)=f''(0)$. The lemma shows that $Qf=Qp$.
answered Jul 23 at 15:53
David C. Ullrich
54.1k33481
add a comment |Â
up vote
1
down vote
accepted
That's curious. I'm going to consider the case $n=1$.
Lemma. If $f(0)=f'(0)=f''(0)=0$ then $Q f=0$.
Proof. For $epsilon>0$ let $$f_epsilon(x)=f(x)+epsilon x^2.$$
Taylor's theorem shows that $f(x)=o(x^2)$ as $xto0$, so for every $epsilon>0$ we have $f_epsilonge0$ in a neighborhood of the origin. Since $f_epsilon(0)=0$ this shows that $$Qf_epsilonge0quad(epsilon>0).$$
So if $g(x)=x^2$ and $alpha=Qg$ we have $$Qf+epsilonalphage0$$for every $epsilon>0$, hence $$Qfge0.$$
But $-f$ satisfies the same hypothesis, so we also have $$-Qfge0,$$hence $Qf=0$.
Outline of how the result follows: Say $P$ is the space of polynomials of degree no larger than $2$. Show that there exist $a,b,c$ so that $$Qp=ap(0)+bp'(0)+cp''(0)quad(pin P).$$
Now for every $f$ there exists $pin P$ with $p(0)=f(0)$, $p'(0)=f'(0)$ and $p''(0)=f''(0)$. The lemma shows that $Qf=Qp$.
answered Jul 23 at 15:53
David C. Ullrich
54.1k33481
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
That's curious. I'm going to consider the case $n=1$.
Lemma. If $f(0)=f'(0)=f''(0)=0$ then $Q f=0$.
Proof. For $epsilon>0$ let $$f_epsilon(x)=f(x)+epsilon x^2.$$
Taylor's theorem shows that $f(x)=o(x^2)$ as $xto0$, so for every $epsilon>0$ we have $f_epsilonge0$ in a neighborhood of the origin. Since $f_epsilon(0)=0$ this shows that $$Qf_epsilonge0quad(epsilon>0).$$
So if $g(x)=x^2$ and $alpha=Qg$ we have $$Qf+epsilonalphage0$$for every $epsilon>0$, hence $$Qfge0.$$
But $-f$ satisfies the same hypothesis, so we also have $$-Qfge0,$$hence $Qf=0$.
Outline of how the result follows: Say $P$ is the space of polynomials of degree no larger than $2$. Show that there exist $a,b,c$ so that $$Qp=ap(0)+bp'(0)+cp''(0)quad(pin P).$$
Now for every $f$ there exists $pin P$ with $p(0)=f(0)$, $p'(0)=f'(0)$ and $p''(0)=f''(0)$. The lemma shows that $Qf=Qp$.
answered Jul 23 at 15:53
David C. Ullrich
54.1k33481
That's curious. I'm going to consider the case $n=1$.
Lemma. If $f(0)=f'(0)=f''(0)=0$ then $Q f=0$.
Proof. For $epsilon>0$ let $$f_epsilon(x)=f(x)+epsilon x^2.$$
Taylor's theorem shows that $f(x)=o(x^2)$ as $xto0$, so for every $epsilon>0$ we have $f_epsilonge0$ in a neighborhood of the origin. Since $f_epsilon(0)=0$ this shows that $$Qf_epsilonge0quad(epsilon>0).$$
So if $g(x)=x^2$ and $alpha=Qg$ we have $$Qf+epsilonalphage0$$for every $epsilon>0$, hence $$Qfge0.$$
But $-f$ satisfies the same hypothesis, so we also have $$-Qfge0,$$hence $Qf=0$.
Outline of how the result follows: Say $P$ is the space of polynomials of degree no larger than $2$. Show that there exist $a,b,c$ so that $$Qp=ap(0)+bp'(0)+cp''(0)quad(pin P).$$
Now for every $f$ there exists $pin P$ with $p(0)=f(0)$, $p'(0)=f'(0)$ and $p''(0)=f''(0)$. The lemma shows that $Qf=Qp$.
answered Jul 23 at 15:53
David C. Ullrich
54.1k33481
edited Jul 23 at 16:07
answered Jul 23 at 15:53
David C. Ullrich
54.1k33481
answered Jul 23 at 15:53
David C. Ullrich
54.1k33481
answered Jul 23 at 15:53
David C. Ullrich
54.1k33481
54.1k33481
add a comment |Â
add a comment |Â
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1
The "Using the linearity of $Q$..." is not right, because $zeta$ is not a scalar! In fact $zeta$ is a function of $x$...
– David C. Ullrich
Jul 23 at 15:58
Ah yes, good point
– Tom Chalmer
Jul 23 at 17:28