Bounded differentiation operator

Question

Is there an example of a normed space $X$, $\dim X=\infty$ such that the differentiation operator $T=\frac{d}{dx}$ is bounded, meaning that $T \in B(X)$ ?

Answer 1

Let $X$ be the set of all functions of the form $$ \frac{1}{\sqrt{2\pi}}\int_{-R}^{R}f(s)e^{-isx}ds, \;\; f \in L^2[-R,R]. $$ This is a linear space over $\mathbb{C}$. You can use the $L^2(\mathbb{R})$ norm on $X$. By the Plancherel Theorem, $$ \left\|\frac{1}{\sqrt{2\pi}}\int_{-R}^{R}f(s)e^{-isx}ds\right\|=\|f\|_{L^2[-R,R]}. $$ This space $X$ is an infinite-dimensional Hilbert space because it is unitarily equivalent to $L^2[-R,R]$.

The functions in $X$ are infinitely differentiable, with \begin{align} \left\|\frac{d}{dx}\frac{1}{\sqrt{2\pi}}\int_{-R}^{R}f(s)e^{-isx}ds\right\| & = \left\|\frac{1}{\sqrt{2\pi}}\int_{-R}^{R}(-is)f(s)e^{-isx}ds\right\| \\ & = \left(\int_{-R}^{R}s^2|f(s)|^2ds\right)^{1/2}\\ & \le R\left(\int_{-R}^{R}|f(s)|^2ds\right)^{1/2}\\ & =R\left\|\frac{1}{\sqrt{2\pi}}\int_{-R}^{R}f(s)e^{-isx}ds\right\| \end{align} So the differentiation operator is bounded on $X$ with a bound of $R$.

Answer 2

Another example: Let $I\subset\mathbb{R}$ be an interval and $X=\{f\in C^\infty(I)\mid \sup_{n\geq 0}\|f^{(n)}\|_\infty<\infty\}$ endowed with the norm $\|f\|_X=\sup_{n\geq 0}\|f^{(n)}\|_\infty$. Then $\|f'\|_X=\sup_{n\geq 1}\|f^{(n)}\|_\infty\leq \|f\|_X$. If $I$ is bounded, then $X$ contains all polynomials, so it's certainly infinite-dimensional.

Of course, this can be done more generally: Let $Y$ be a normed space and $T\colon D(T)\to Y$ a linear operator defined on a subspace $D(T)\subset Y$. Let $X=\{u\in\bigcap_{n\geq 0} D(T^n)\mid \sup_{n\geq 0}\|T^n u\|_Y<\infty\}$ with norm $\|u\|_X=\sup_{n\geq 0}\|T^n u\|_Y$. Then $T\colon X\to X$ is bounded. If $X$ is infinite-dimensional depends on the operator $T$.

Remark: As usual, $D(T^n)$ is defined recursively by $D(T^{n+1})=\{u\in D(T)\mid Tu\in D(T^n)\}$.