# Is this linear operator on polynomials with sup-norm bounded?

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!

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 $$\ell_{2n-1}(p_m)\xrightarrow[m\to\infty]{}\infty$$.

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

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 !

• Contrariwise, I understood the defintion to be $$\ell_n(a_mx^m + a_{m-1}x^{m-1} + \dots + a_1x +a_0) = a_n$$ – GEdgar Jan 30 '19 at 1:50
• @GEdgar Yes, $\ell_n$ is defined as this – zzz Jan 30 '19 at 2:48
• -1: I fail to see how this can possibly lead to an answer. I'll be happy to be proven wrong. – Martin Argerami Jan 30 '19 at 18:00