# Problem 3, chapter 2 Rudin functional analysis

Put $$K = [-1,1]$$; define $$\mathcal{D}_K$$ as in section 1.46 with ($$\mathbb{R}$$ in place of $$\mathbb{R}^n$$). Suppose $$\left\{ f_n > \right\}$$ is a sequence of Lebesgue integrable functions such that

$$\Lambda \phi = \lim_{n,\infty} \int_{-1}^1 f_n(t)\phi(t)dt$$ exists for every $$\phi \in \mathcal{D}_K$$. Show that $$\Lambda$$ is a continuous linear functional on $$\mathcal{D}_K$$. Show that there's a positive integer $$p$$ and $$M < \infty$$ such that

$$\left| \int_{-1}^1 f_n(t)\phi(t) dt \right| \leq M \lVert D^p \phi \rVert_\infty$$

Just a note $$\mathcal{D}_K$$ is the set of all functions in $$C^\infty(K)$$ whose supports is in $$K$$.

I really don't have a clue for the existance of $$M$$ and $$p$$. For the first part instead I defined

$$\Lambda_n \phi = \int_{-1}^1 f_n(t)\phi(t)dt$$

And from here I thought I could use the following theorem, which follows more or less from the Banach-Steinhaus theorem

Theorem 2.8 If $$\left\{ \Lambda_n \right\}$$ is a sequence of continuous linear mappings from a F-space $$X$$ into a topological vector space $$Y$$, and if $$\Lambda x = \lim_{n,\infty} \Lambda_n x$$ exists for every $$x$$ in $$X$$, then $$\Lambda$$ is continuous

Since each $$\Lambda_n : \mathcal{D}_K \to \mathbb{R}$$, and $$\mathcal{D}_K$$ is a Frechet space, and hence an $$F-space$$ the only bit to be proven in order to apply theorem 2.8 I need to prove the continuity I'd use

Theorem 1.18

Let $$\Lambda$$ a linear functionals on a topological vector space $$X$$. Assume $$\Lambda x \neq 0$$ for some $$x \in X$$. Then each of the following four properties implies the other three

a) $$\Lambda$$ is continuous

b) The null space $$\mathcal{N}(\Lambda)$$ is closed

c) $$\mathcal{N}(\Lambda)$$ is not dense in $$X$$

d) $$\Lambda$$ is bounded in some neighborhood $$V$$ of $$0$$.

Here is where I start to get confused... If I pick any neighborhood of $$0$$, call such neighborhood $$V$$ since $$\Lambda_n$$ exist for every $$\phi$$ in $$\mathcal{D}_K$$ the abs value of the integral is bounded on $$V$$.

This implies that d) of theorem 1.18 is true, hence $$\Lambda_n$$ is continuous and by theorem 2.8 $$\Lambda$$ is continuous.

Is this argument correct?

As said instead for the second part I have no clue at all and any hint would be highly appreciated.

Update

I'm probably missing, among the things, that $$M$$ and $$p$$ are indepedent from $$n$$ but they might change with $$\phi$$.

Update 2

I've just found this solution, the only bit I'm not entirely convinced is the family of seminorms used, they look different from the one defined in section 1.46 of the book. Al the rest seems to make sense, but I'd still like a confirmation.

• The stated inequality precisely means that $\Lambda_n$ are equicontinuous on $\mathscr D_{[-1,1]}$. Apply Banach-Steinhaus. – Jochen May 20 '19 at 7:00
• What about the first part? – user8469759 May 20 '19 at 8:09
• Also isn't the equicontinuous a consequence that each $\Lambda_n$ is a continuous linear functional? (Still using theorem 2.8). This means that each $\Lambda_n$ is bounded, and hence the equicontinuity (there's also the second category to be proved, but this follows from the fact that $\mathscr D_K$ is a frechet space. – user8469759 May 20 '19 at 9:43
• I mean you use theorem 2.8 which enables to use banach-steinhaus, and hence equicontinuity. However I'm not sure where the $D^p \phi$ comes from, to me this is a consequence of the topology defined on $\mathscr D_K$ – user8469759 May 20 '19 at 9:44

From the proof of Theorem 2.8 you get that the family $$\{\Lambda_{n}\}_{n}$$ is equi-continuous. Hence, given $$\varepsilon=1$$ there exists an open neighborhood $$W$$ of the origin such that $$\Lambda_{n}(W)\subseteq(-1,1)$$ for all $$n$$. But since the topology of $$\mathcal{D}_{K}$$ is generated by the norms $$\Vert\cdot\Vert_{p}$$ you can find $$r>0$$ and $$p\in\mathbb{N}_{0}$$ such that $$B_{p}(0,r)\subseteq W$$, and so $$\Lambda_{n}(B_{p}(0,r))\subseteq(-1,1)$$ for all $$n$$, that is, $$|\Lambda_{n}(\phi)|<1$$ for all $$\phi\in\mathcal{D}_{K}$$ with $$\Vert\phi\Vert_{p}. If $$\phi \in\mathcal{D}_{K}$$ and $$\phi\neq0$$, then $$\Vert\phi\Vert_{p}\neq0$$ and $$\Bigl\Vert r\frac{\phi}{2\Vert\phi\Vert_{p}}\Bigr\Vert_{p} By the linearity of $$\Lambda_{n}$$ it follows that $$|\Lambda_{n}(\phi )|\leq2r^{-1}\Vert\phi\Vert_{p}$$. Now, $$\Vert\phi\Vert_{p}=\max\{|D^{n}\phi(x)|:\,x\in K,\,n=0,\ldots,p\}.$$ Using Taylor's formula with center at $$-1$$ and integral remainder for $$\phi$$ and all its derivatives of order less than $$p$$ you get \begin{align*} D^{n}\phi(x) & =D^{n}\phi(-1)+\sum_{k=1}^{p-n-1}\frac{1}{k!}D^{n+k}% \phi(-1)(x+1)^{k}+\frac{1}{(p-n)!}\int_{-1}^{x}D^{p}\phi(t)(x-t)^{p-n}dt\\ & =0+0+\frac{1}{(p-n)!}\int_{-1}^{x}D^{p}\phi(t)(x-t)^{p-n}dt \end{align*} and so $$|D^{n}\phi(x)|\leq\frac{\Vert D^{p}\phi\Vert_{\infty}}{(p-n)!}\int_{-1}% ^{1}|x-t|^{p-n}dt$$ for all $$x\in\lbrack-1,1]$$ and all $$n=0,\ldots,p-1$$. Hence, $$\Vert\phi\Vert_{p}\leq C\Vert D^{p}\phi\Vert_{\infty}%$$ It follows that$$|\Lambda_{n}(\phi )|\leq2r^{-1}\Vert\phi\Vert_{p}\le 2r^{-1}C\Vert D^{p}\phi\Vert_{\infty}$$
• I am using the seminorms defined on page 33 formula 4. Since you have functions of one variable the multi-indeces are just integers and I am taking $K_1=K$. They are seminorms in $C^\infty(\mathbb{R})$ but they are norms in the space of smooth functions with support in $K$. If you want, just write $q_p(\cdot)$ instead of $\|\cdot\|_p$. – Gio67 Dec 18 '19 at 21:22
• I'm reading through your proof. First of all I just want to make sure I understand, it might take a bit before I accept the answer, the $p$ is independent from both $n$ and $\phi$, right? – user8469759 Dec 19 '19 at 10:31
• yes. The set $W$ does not depend on $n$ (and there is no $\phi$) – Gio67 Dec 19 '19 at 15:02
• Are you applying the open mapping theorem? (specifically When you pick the ball $B_p(0,r)$)? Or are you just using theorem 2.4? – user8469759 Dec 20 '19 at 12:28