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I'm working on the following question in Royden:

Let $X$ be the linear space of all polynomials defined on $\mathbb{R}$. For $p \in X$, define $\|p\|$ to be the sums of the absolute values of the coefficients of $p$. Show that this is a norm on $X$. For each $n$ define $\psi_n: X \to \mathbb{R}$ by $\psi_n(p)=p^{(n)}(0)$. Use the properties of the sequence $\psi_n$ in $L(X,\mathbb{R})$ to show that $X$ is not a Banach Space.

Showing the norm part is straight forward. I want to say something like -- there is a pointwise bound but no uniform bound, so by Banach-Steinhaus, $X$ is not complete. Here's my half-solution:

We can establish pointwise bounds like so: $$\psi_n(p) \leq c_0 \text{ (the constant coeffcient in the polynomial)} \leq \|p\|$$

To get a failure of uniform bound I want to choose a particularly bad $p \in X$ so that there's an $n$ in the LHS of the last inequality: $$|\psi_n(p)| \leq \|p\| \|\psi_n\| \implies \frac{|\psi_n(p)|}{\|p\|} \leq \|\psi_n\| \implies \frac{|p^{(n)}(0)|}{\|p\|} \leq \|\psi_n\|$$

Do people see what might work? Or do I need a different argument?

There a similar question here, but it didn't help me.

yoshi
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2 Answers2

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For any $p$, the sequence $\psi_n(p)$ is constantly equal to $0$ as soon as $n$ is greater than the degree of $p$. In particular, $\sup_n(|\psi_n(p)|) < \infty$.

On the other hand, $||\psi_n|| \geq n!$ since $\frac{|\psi_n(x^n)|}{|x^n|} = n!$.

It follows that $X$ does not satisfy the conclusion of the Banach-Steinhaus theorem and thus cannot be a Banach space.

hunter
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  • Degree doesn't matter for $\sup_n(|\psi_n(p)|)$ right? Evaluation at zero for any polynomial of arbitrary degree will kill all the values except the constant term in the polynomial. 2. Doesn't $|\psi_n(x^n)| = 0^n = 0$? 3. In general, how are we interpreting $\psi_n$ - I take it to mean truncate polynomials to degree $n$, then evaluate this at zero. I'm not sure if this is right though.
    • – yoshi Apr 18 '18 at 03:44
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    @yoshi I interpret $\psi_n$ as taking $n$ derivatives, then evaluating at $0$, which I believe answers both your questions. – hunter Apr 18 '18 at 04:03
  • okay, I think I like your interpretation better. – yoshi Apr 18 '18 at 04:04