Continuity in Frechet spaces

Question

These are undoubtably simple questions, but I have no background in functional analysis and am wondering about them. The first is an exercise from Folland, the second is not, but both are claims I've seen in a paper I'm reading.

Consider the space $C^\infty(\mathbb R)$ of all infinitely differentiable functions on $\mathbb R$. Why does $f_n\rightarrow f$ if and only if $f^{k}_n \rightarrow f^{k}$ uniformly on compact sets for all $k\ge 0$?

Consider the space $C^\infty(\mathbb T)$ of $2\pi$-periodic functions. Why is the operation $(f,g)\rightarrow fg$ (pointwise multiplication) continuous in this topology?

Answer 1

When $E$ is a vector space and $(p_i)_{i\in I}$ is a family of semi-norms, we say that the sequence $\{x_n\}$ converges to $x$ if and only if $p_i(x_n-x)\to 0$ for each $i\in I$.

• When we work with the space of smooth functions defined on the real line, we can take $I:=\Bbb N\times\Bbb N$ and $p_{n,d}(f):=\sup_{|x|\leq n}|f^{(d)}(x)|$. To see the equivalence, just see that the compact sets are bounded for one direction, and for the other choose particular compact sets.

• When we deal with functions of $C^{\infty}(\mathbb T)$, the semi-norms $p_d(f):=\max_{1\leq k\leq }d\sup_{x\in\mathbb T}|f^{(d)}(x)|$ are natural, as the torus is compact. The map $(f,g)\mapsto fg$ is bilinear. We have $$p_d(fg)\leq \max_{1\leq n\leq d}\sup_{x\in\Bbb T}\sum_{k=0}^n\binom nk|f^{(k)}(x)||g^{(n-k)}(x)|\leq \max_{1\leq n\leq d}2^np_n(f)p_n(g)\leq 2^dp_d(f)p_d(g),$$ which ensures continuity for the topology induced by the semi-norms.

Answer 2

For the begining I recall some facts from the theory of general locally convex spaces.

Let $X$ be a linear space with family of seminorms $\{\Vert\cdot\Vert_\lambda:\lambda\in\Lambda\}$. Then $X$ can be made a locally convex topological space. Pre-base of this locally convex topology $\tau$ is given by family of sets $$ B_{x_0,\varepsilon,\lambda}=\{x\in X:\Vert x-x_0\Vert_\lambda<\varepsilon\},\quad\text{where}\quad x_0\in X,\;\varepsilon>0,\;\lambda\in L $$ In fact topology of each locally convex space can be described by some family of seminorms, and moreover most of definitions in the theory of locally convex can be translated into the language of seminorms.

For example, let $(X,\{\Vert\cdot\Vert_\lambda:\lambda\in L\})$, $(Y,\{\Vert\cdot\Vert_\mu:\mu\in M\})$ and $(Z,\{\Vert\cdot\Vert_\nu:\nu\in N\})$ be locally convex spaces, then bilinear operator $T:X\times Y\to Z$ is continuous iff
$$ \forall\nu\in N\quad\exists C>0\quad \exists\lambda_1,\ldots,\lambda_n\in L\quad\exists\mu_1,\ldots,\mu_m\in M\quad $$ $$ \Vert T(x,y)\Vert_\nu\leq C\max\limits_{i=1,\ldots,n}\Vert x\Vert_{\lambda_i}\max\limits_{i=1,\ldots,m}\Vert y\Vert_{\mu_i} $$

Another example: convergence of sequences in $X$ can be described in terms of seminorms $$ \lim\limits_{n\to\infty} x_n\underset{\tau}{=}x \quad\Longleftrightarrow\quad \forall\lambda\in\Lambda\quad\lim\limits_{n\to\infty}\Vert x_n-x\Vert_\lambda=0 $$

As for the first question, in your situation $$ X=C^\infty(\mathbb{R}) $$ $$ L=\{(K,i)\subset \mathbb{R}:\; K\text{ - is compact},\; i\in\mathbb{N}_0\} $$ $$ \Vert x\Vert_\lambda=\sup\{|f^{(i)}(t)|:t\in K\},\qquad\lambda=(K,i) $$ Now it is remains to recall that convergence in $\sup$ norm is uniform convergence, and the result follows.

As for the second question, in your situation $$ X=Y=Z=C^\infty(\mathbb{T}) $$ $$ L=M=N=\{(K,n)\subset \mathbb{R}:\; K\text{ - is compact},\; n\in\mathbb{N}_0\} $$ $$ T:C^\infty(\mathbb{T})\times C^\infty(\mathbb{T})\to C^\infty(\mathbb{T}):(x,y)\mapsto xy $$ Now take arbitrary $\nu=(K,n)\in N$ and consider $C=2^n$ and $\lambda_i=\mu_i=(K,i)$ with $i=1,\ldots,n$, then $$ \begin{align} \Vert T(x,y)\Vert_\nu &=\sup\{|(xy)^{(n)}(t)|:t\in K\}\\ &=\sup\left\{\left|\sum\limits_{i=0}^n {n\choose i}x^{(i)}(t)y^{(n-i)}(t)\right|:t\in K\right\}\\ &\leq\sup\left\{\sum\limits_{i=0}^n {n\choose i}|x^{(i)}(t)||y^{(n-i)}(t)|:t\in K\right\}\\ &\leq\sum\limits_{i=0}^n {n\choose i}\sup\{|x^{(i)}(t)||y^{(n-i)}(t)|:t\in K\}\\ &\leq\sum\limits_{i=0}^n {n\choose i}\sup\{|x^{(i)}(t)|:t\in K\}\sup\{|y^{(n-i)}(t)|:t\in K\}\\ &\leq\sum\limits_{i=0}^n {n\choose i}\Vert x\Vert_{(K,i)}\Vert y\Vert_{(K,n-i)}\\ &\leq 2^n\max\limits_{i=1,\ldots,n}\Vert x\Vert_{\lambda_i}\max\limits_{i=1,\ldots,n}\Vert y\Vert_{\mu_i} \end{align} $$ Hence $T$ is continuous.