Mixing
Is this linear operator on polynomials with sup-norm bounded?
$begingroup$
Question: Let $mathcal{P}$ be the space of all polynomials (with real coefficients) on the real line, endowed with sup-norm (i.e., $|p| = sup_{0le xle 1}|p(x)|$).
For any fixed $n in mathbb{N}$, consider the linear functional $ell_n colon mathcal{P} to mathbb{R}$, where $ell_n(p)$
is equal to the coefficient of $x^n$ in $p$. Is $ell_n$ a bounded linear functional on this normed (but incomplete) space?
Attempt: Well I can see that $ell_0$ is a bounded linear functional with norm 1, but I don't know the answer in general. Thanks for your help!
functional-analysis normed-spaces
asked Jan 30 at 1:37
zzzzzz
374
$endgroup$
add a comment |
$begingroup$
Question: Let $mathcal{P}$ be the space of all polynomials (with real coefficients) on the real line, endowed with sup-norm (i.e., $|p| = sup_{0le xle 1}|p(x)|$).
For any fixed $n in mathbb{N}$, consider the linear functional $ell_n colon mathcal{P} to mathbb{R}$, where $ell_n(p)$
is equal to the coefficient of $x^n$ in $p$. Is $ell_n$ a bounded linear functional on this normed (but incomplete) space?
Attempt: Well I can see that $ell_0$ is a bounded linear functional with norm 1, but I don't know the answer in general. Thanks for your help!
functional-analysis normed-spaces
asked Jan 30 at 1:37
zzzzzz
374
$endgroup$
add a comment |
$begingroup$
Question: Let $mathcal{P}$ be the space of all polynomials (with real coefficients) on the real line, endowed with sup-norm (i.e., $|p| = sup_{0le xle 1}|p(x)|$).
For any fixed $n in mathbb{N}$, consider the linear functional $ell_n colon mathcal{P} to mathbb{R}$, where $ell_n(p)$
is equal to the coefficient of $x^n$ in $p$. Is $ell_n$ a bounded linear functional on this normed (but incomplete) space?
Attempt: Well I can see that $ell_0$ is a bounded linear functional with norm 1, but I don't know the answer in general. Thanks for your help!
functional-analysis normed-spaces
asked Jan 30 at 1:37
zzzzzz
374
$endgroup$
Question: Let $mathcal{P}$ be the space of all polynomials (with real coefficients) on the real line, endowed with sup-norm (i.e., $|p| = sup_{0le xle 1}|p(x)|$).
For any fixed $n in mathbb{N}$, consider the linear functional $ell_n colon mathcal{P} to mathbb{R}$, where $ell_n(p)$
is equal to the coefficient of $x^n$ in $p$. Is $ell_n$ a bounded linear functional on this normed (but incomplete) space?
Attempt: Well I can see that $ell_0$ is a bounded linear functional with norm 1, but I don't know the answer in general. Thanks for your help!
functional-analysis normed-spaces
functional-analysis normed-spaces
asked Jan 30 at 1:37
zzzzzz
374
asked Jan 30 at 1:37
zzzzzz
374
asked Jan 30 at 1:37
zzzzzz
374
asked Jan 30 at 1:37
zzzzzz
374
asked Jan 30 at 1:37
zzzzzz
374
374
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
The map $ell_n$ is not bounded for $ngeq1$. I will do the case $n$ odd, but the even case can be done with the same idea.
Consider the function $f_m(x)=tfrac1m,sin(m^2 x)$. It's easy to check that $|f_m|=1/m$ and that
$$
f_m^{(2n-1)}(0)=m^{4n-3}.
$$
Let $p_m$ be the Taylor polynomial of $f_m$, of degree big enough so that $|f_m-p_m|<1/m$ on $[0,1]$. Then
$$
|p_m|leq|p_m-f_m|+|f_m|leqfrac2m,
$$
while
$$
ell_{2n-1}(p_m)=frac{m^{4n-3}}{(2n-1)!}.
$$
As we can do this for all $minmathbb N$, we obtain a sequence ${p_m}$ with $|p_m|to0$ and $ell_{2n-1}(p_m)xrightarrow[mtoinfty]{}infty$.
answered Jan 30 at 2:05
Martin ArgeramiMartin Argerami
129k1184185
$endgroup$
$begingroup$
Well, I don't see why ‖p‖ is what you claimed. ‖p‖ must be at least p(1) = 1-c, so if you choose c to be small, won't ‖p‖ be very close to 1?
$endgroup$
– zzz
Jan 30 at 2:46
$begingroup$
You want $c$ to be big, but you are right, my example doesn't work.
$endgroup$
– Martin Argerami
Jan 30 at 2:47
$begingroup$
Please check the new version.
$endgroup$
– Martin Argerami
Jan 30 at 3:38
$begingroup$
Thanks! This is a nice counterexample.
$endgroup$
– zzz
Jan 30 at 5:35
add a comment |
$begingroup$
Hint :
Express your functional as :
$$ell_n(x) = a_nx^n + a_{n-1}x^{n-1} + dots + a_1x +a_0$$
Now, take $|ell_n(x)|$ and try to form an inequality involving the sup norm. Triangle inequality will be your friend !
answered Jan 30 at 1:43
RebellosRebellos
15.6k31250
$endgroup$
2
$begingroup$
Contrariwise, I understood the defintion to be $$ ell_n(a_mx^m + a_{m-1}x^{m-1} + dots + a_1x +a_0) = a_n $$
$endgroup$
– GEdgar
Jan 30 at 1:50
$begingroup$
@GEdgar Yes, $ell_n$ is defined as this
$endgroup$
– zzz
Jan 30 at 2:48
$begingroup$
-1: I fail to see how this can possibly lead to an answer. I'll be happy to be proven wrong.
$endgroup$
– Martin Argerami
Jan 30 at 18:00
add a comment |
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2 Answers
2
active
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2 Answers
2
active
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$begingroup$
The map $ell_n$ is not bounded for $ngeq1$. I will do the case $n$ odd, but the even case can be done with the same idea.
Consider the function $f_m(x)=tfrac1m,sin(m^2 x)$. It's easy to check that $|f_m|=1/m$ and that
$$
f_m^{(2n-1)}(0)=m^{4n-3}.
$$
Let $p_m$ be the Taylor polynomial of $f_m$, of degree big enough so that $|f_m-p_m|<1/m$ on $[0,1]$. Then
$$
|p_m|leq|p_m-f_m|+|f_m|leqfrac2m,
$$
while
$$
ell_{2n-1}(p_m)=frac{m^{4n-3}}{(2n-1)!}.
$$
As we can do this for all $minmathbb N$, we obtain a sequence ${p_m}$ with $|p_m|to0$ and $ell_{2n-1}(p_m)xrightarrow[mtoinfty]{}infty$.
answered Jan 30 at 2:05
Martin ArgeramiMartin Argerami
129k1184185
$endgroup$
$begingroup$
Well, I don't see why ‖p‖ is what you claimed. ‖p‖ must be at least p(1) = 1-c, so if you choose c to be small, won't ‖p‖ be very close to 1?
$endgroup$
– zzz
Jan 30 at 2:46
$begingroup$
You want $c$ to be big, but you are right, my example doesn't work.
$endgroup$
– Martin Argerami
Jan 30 at 2:47
$begingroup$
Please check the new version.
$endgroup$
– Martin Argerami
Jan 30 at 3:38
$begingroup$
Thanks! This is a nice counterexample.
$endgroup$
– zzz
Jan 30 at 5:35
add a comment |
$begingroup$
The map $ell_n$ is not bounded for $ngeq1$. I will do the case $n$ odd, but the even case can be done with the same idea.
Consider the function $f_m(x)=tfrac1m,sin(m^2 x)$. It's easy to check that $|f_m|=1/m$ and that
$$
f_m^{(2n-1)}(0)=m^{4n-3}.
$$
Let $p_m$ be the Taylor polynomial of $f_m$, of degree big enough so that $|f_m-p_m|<1/m$ on $[0,1]$. Then
$$
|p_m|leq|p_m-f_m|+|f_m|leqfrac2m,
$$
while
$$
ell_{2n-1}(p_m)=frac{m^{4n-3}}{(2n-1)!}.
$$
As we can do this for all $minmathbb N$, we obtain a sequence ${p_m}$ with $|p_m|to0$ and $ell_{2n-1}(p_m)xrightarrow[mtoinfty]{}infty$.
answered Jan 30 at 2:05
Martin ArgeramiMartin Argerami
129k1184185
$endgroup$
$begingroup$
Well, I don't see why ‖p‖ is what you claimed. ‖p‖ must be at least p(1) = 1-c, so if you choose c to be small, won't ‖p‖ be very close to 1?
$endgroup$
– zzz
Jan 30 at 2:46
$begingroup$
You want $c$ to be big, but you are right, my example doesn't work.
$endgroup$
– Martin Argerami
Jan 30 at 2:47
$begingroup$
Please check the new version.
$endgroup$
– Martin Argerami
Jan 30 at 3:38
$begingroup$
Thanks! This is a nice counterexample.
$endgroup$
– zzz
Jan 30 at 5:35
add a comment |
$begingroup$
The map $ell_n$ is not bounded for $ngeq1$. I will do the case $n$ odd, but the even case can be done with the same idea.
Consider the function $f_m(x)=tfrac1m,sin(m^2 x)$. It's easy to check that $|f_m|=1/m$ and that
$$
f_m^{(2n-1)}(0)=m^{4n-3}.
$$
Let $p_m$ be the Taylor polynomial of $f_m$, of degree big enough so that $|f_m-p_m|<1/m$ on $[0,1]$. Then
$$
|p_m|leq|p_m-f_m|+|f_m|leqfrac2m,
$$
while
$$
ell_{2n-1}(p_m)=frac{m^{4n-3}}{(2n-1)!}.
$$
As we can do this for all $minmathbb N$, we obtain a sequence ${p_m}$ with $|p_m|to0$ and $ell_{2n-1}(p_m)xrightarrow[mtoinfty]{}infty$.
answered Jan 30 at 2:05
Martin ArgeramiMartin Argerami
129k1184185
$endgroup$
The map $ell_n$ is not bounded for $ngeq1$. I will do the case $n$ odd, but the even case can be done with the same idea.
Consider the function $f_m(x)=tfrac1m,sin(m^2 x)$. It's easy to check that $|f_m|=1/m$ and that
$$
f_m^{(2n-1)}(0)=m^{4n-3}.
$$
Let $p_m$ be the Taylor polynomial of $f_m$, of degree big enough so that $|f_m-p_m|<1/m$ on $[0,1]$. Then
$$
|p_m|leq|p_m-f_m|+|f_m|leqfrac2m,
$$
while
$$
ell_{2n-1}(p_m)=frac{m^{4n-3}}{(2n-1)!}.
$$
As we can do this for all $minmathbb N$, we obtain a sequence ${p_m}$ with $|p_m|to0$ and $ell_{2n-1}(p_m)xrightarrow[mtoinfty]{}infty$.
answered Jan 30 at 2:05
Martin ArgeramiMartin Argerami
129k1184185
edited Jan 30 at 3:37
answered Jan 30 at 2:05
Martin ArgeramiMartin Argerami
129k1184185
answered Jan 30 at 2:05
Martin ArgeramiMartin Argerami
129k1184185
answered Jan 30 at 2:05
Martin ArgeramiMartin Argerami
129k1184185
129k1184185
$begingroup$
Well, I don't see why ‖p‖ is what you claimed. ‖p‖ must be at least p(1) = 1-c, so if you choose c to be small, won't ‖p‖ be very close to 1?
$endgroup$
– zzz
Jan 30 at 2:46
$begingroup$
You want $c$ to be big, but you are right, my example doesn't work.
$endgroup$
– Martin Argerami
Jan 30 at 2:47
$begingroup$
Please check the new version.
$endgroup$
– Martin Argerami
Jan 30 at 3:38
$begingroup$
Thanks! This is a nice counterexample.
$endgroup$
– zzz
Jan 30 at 5:35
add a comment |
$begingroup$
Well, I don't see why ‖p‖ is what you claimed. ‖p‖ must be at least p(1) = 1-c, so if you choose c to be small, won't ‖p‖ be very close to 1?
$endgroup$
– zzz
Jan 30 at 2:46
$begingroup$
You want $c$ to be big, but you are right, my example doesn't work.
$endgroup$
– Martin Argerami
Jan 30 at 2:47
$begingroup$
Please check the new version.
$endgroup$
– Martin Argerami
Jan 30 at 3:38
$begingroup$
Thanks! This is a nice counterexample.
$endgroup$
– zzz
Jan 30 at 5:35
$begingroup$
Well, I don't see why ‖p‖ is what you claimed. ‖p‖ must be at least p(1) = 1-c, so if you choose c to be small, won't ‖p‖ be very close to 1?
$endgroup$
– zzz
Jan 30 at 2:46
$begingroup$
Well, I don't see why ‖p‖ is what you claimed. ‖p‖ must be at least p(1) = 1-c, so if you choose c to be small, won't ‖p‖ be very close to 1?
$endgroup$
– zzz
Jan 30 at 2:46
$begingroup$
You want $c$ to be big, but you are right, my example doesn't work.
$endgroup$
– Martin Argerami
Jan 30 at 2:47
$begingroup$
You want $c$ to be big, but you are right, my example doesn't work.
$endgroup$
– Martin Argerami
Jan 30 at 2:47
$begingroup$
Please check the new version.
$endgroup$
– Martin Argerami
Jan 30 at 3:38
$begingroup$
Please check the new version.
$endgroup$
– Martin Argerami
Jan 30 at 3:38
$begingroup$
Thanks! This is a nice counterexample.
$endgroup$
– zzz
Jan 30 at 5:35
$begingroup$
Thanks! This is a nice counterexample.
$endgroup$
– zzz
Jan 30 at 5:35
add a comment |
$begingroup$
Hint :
Express your functional as :
$$ell_n(x) = a_nx^n + a_{n-1}x^{n-1} + dots + a_1x +a_0$$
Now, take $|ell_n(x)|$ and try to form an inequality involving the sup norm. Triangle inequality will be your friend !
answered Jan 30 at 1:43
RebellosRebellos
15.6k31250
$endgroup$
2
$begingroup$
Contrariwise, I understood the defintion to be $$ ell_n(a_mx^m + a_{m-1}x^{m-1} + dots + a_1x +a_0) = a_n $$
$endgroup$
– GEdgar
Jan 30 at 1:50
$begingroup$
@GEdgar Yes, $ell_n$ is defined as this
$endgroup$
– zzz
Jan 30 at 2:48
$begingroup$
-1: I fail to see how this can possibly lead to an answer. I'll be happy to be proven wrong.
$endgroup$
– Martin Argerami
Jan 30 at 18:00
add a comment |
$begingroup$
Hint :
Express your functional as :
$$ell_n(x) = a_nx^n + a_{n-1}x^{n-1} + dots + a_1x +a_0$$
Now, take $|ell_n(x)|$ and try to form an inequality involving the sup norm. Triangle inequality will be your friend !
answered Jan 30 at 1:43
RebellosRebellos
15.6k31250
$endgroup$
2
$begingroup$
Contrariwise, I understood the defintion to be $$ ell_n(a_mx^m + a_{m-1}x^{m-1} + dots + a_1x +a_0) = a_n $$
$endgroup$
– GEdgar
Jan 30 at 1:50
$begingroup$
@GEdgar Yes, $ell_n$ is defined as this
$endgroup$
– zzz
Jan 30 at 2:48
$begingroup$
-1: I fail to see how this can possibly lead to an answer. I'll be happy to be proven wrong.
$endgroup$
– Martin Argerami
Jan 30 at 18:00
add a comment |
$begingroup$
Hint :
Express your functional as :
$$ell_n(x) = a_nx^n + a_{n-1}x^{n-1} + dots + a_1x +a_0$$
Now, take $|ell_n(x)|$ and try to form an inequality involving the sup norm. Triangle inequality will be your friend !
answered Jan 30 at 1:43
RebellosRebellos
15.6k31250
$endgroup$
Hint :
Express your functional as :
$$ell_n(x) = a_nx^n + a_{n-1}x^{n-1} + dots + a_1x +a_0$$
Now, take $|ell_n(x)|$ and try to form an inequality involving the sup norm. Triangle inequality will be your friend !
answered Jan 30 at 1:43
RebellosRebellos
15.6k31250
answered Jan 30 at 1:43
RebellosRebellos
15.6k31250
answered Jan 30 at 1:43
RebellosRebellos
15.6k31250
answered Jan 30 at 1:43
RebellosRebellos
15.6k31250
15.6k31250
2
$begingroup$
Contrariwise, I understood the defintion to be $$ ell_n(a_mx^m + a_{m-1}x^{m-1} + dots + a_1x +a_0) = a_n $$
$endgroup$
– GEdgar
Jan 30 at 1:50
$begingroup$
@GEdgar Yes, $ell_n$ is defined as this
$endgroup$
– zzz
Jan 30 at 2:48
$begingroup$
-1: I fail to see how this can possibly lead to an answer. I'll be happy to be proven wrong.
$endgroup$
– Martin Argerami
Jan 30 at 18:00
add a comment |
2
$begingroup$
Contrariwise, I understood the defintion to be $$ ell_n(a_mx^m + a_{m-1}x^{m-1} + dots + a_1x +a_0) = a_n $$
$endgroup$
– GEdgar
Jan 30 at 1:50
$begingroup$
@GEdgar Yes, $ell_n$ is defined as this
$endgroup$
– zzz
Jan 30 at 2:48
$begingroup$
-1: I fail to see how this can possibly lead to an answer. I'll be happy to be proven wrong.
$endgroup$
– Martin Argerami
Jan 30 at 18:00
2
2
$begingroup$
Contrariwise, I understood the defintion to be $$ ell_n(a_mx^m + a_{m-1}x^{m-1} + dots + a_1x +a_0) = a_n $$
$endgroup$
– GEdgar
Jan 30 at 1:50
$begingroup$
Contrariwise, I understood the defintion to be $$ ell_n(a_mx^m + a_{m-1}x^{m-1} + dots + a_1x +a_0) = a_n $$
$endgroup$
– GEdgar
Jan 30 at 1:50
$begingroup$
@GEdgar Yes, $ell_n$ is defined as this
$endgroup$
– zzz
Jan 30 at 2:48
$begingroup$
@GEdgar Yes, $ell_n$ is defined as this
$endgroup$
– zzz
Jan 30 at 2:48
$begingroup$
-1: I fail to see how this can possibly lead to an answer. I'll be happy to be proven wrong.
$endgroup$
– Martin Argerami
Jan 30 at 18:00
$begingroup$
-1: I fail to see how this can possibly lead to an answer. I'll be happy to be proven wrong.
$endgroup$
– Martin Argerami
Jan 30 at 18:00
add a comment |
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