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Question: Let $\mathcal{P}$ be the space of all polynomials (with real coefficients) on the real line, endowed with sup-norm (i.e., $\|p\| = \sup_{0\le x\le 1}|p(x)|$).

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!

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The map $\ell_n$ is not bounded for $n\geq1$. 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 $$ \|p_m\|\leq\|p_m-f_m\|+\|f_m\|\leq\frac2m, $$ while $$ \ell_{2n-1}(p_m)=\frac{m^{4n-3}}{(2n-1)!}. $$ As we can do this for all $m\in\mathbb N$, we obtain a sequence $\{p_m\}$ with $\|p_m\|\to0$ and $\ell_{2n-1}(p_m)\xrightarrow[m\to\infty]{}\infty$.

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  • $\begingroup$ 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 '19 at 2:46
  • $\begingroup$ You want $c$ to be big, but you are right, my example doesn't work. $\endgroup$ – Martin Argerami Jan 30 '19 at 2:47
  • $\begingroup$ Please check the new version. $\endgroup$ – Martin Argerami Jan 30 '19 at 3:38
  • $\begingroup$ Thanks! This is a nice counterexample. $\endgroup$ – zzz Jan 30 '19 at 5:35
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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 !

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    $\begingroup$ 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 '19 at 1:50
  • $\begingroup$ @GEdgar Yes, $\ell_n$ is defined as this $\endgroup$ – zzz Jan 30 '19 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 '19 at 18:00

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