Unbounded differential operator in $C^1.$

Question

Why is the operator $\dfrac{d}{dx}$ unbounded? Let us take $f\in C^1[a,b]$ and the operator norm $$\left\|\frac{d}{dx}\right\|=\sup\left(\left\|\frac{d}{dx}f(x)\right\|:\|f(x)\|\leq 1\right).$$

I want some examples for unbounded operators of this form. Doesn't it mean that there are bounded functions with unbounded first derivative? But then, it cannot be $C^1[a,b]$ in the first place.

For $C^1$ use the norm $$||f||=\sup|f(x)|+\sup|f'(x)|$$ on $[a,b]$

Answer 1

I'm assuming that, for you, $\frac{\mathrm d}{\mathrm dx}$ is a map from $C^1\bigl([a,b]\bigr)$ into $C\bigl([a,b]\bigr)$, both of them endowed with the $\sup$ norm. Then, indeed, $\frac{\mathrm d}{\mathrm dx}$ is discontinuous. For instance, all functions $x\mapsto x^n$ ($n\in\mathbb N$) belong to the closide unit ball, but the sequence $\left(\frac{\mathrm d}{\mathrm dx}x^n\right)_{n\in\mathbb N}$ in unbounded.

Besides, there are bounded functions with unbounded first derivative. Take, for instance$$\begin{array}{ccc}[0,1]&\longrightarrow&\mathbb{R}\\x&\mapsto&\begin{cases}x^2\sin\left(\frac1{x^2}\right)&\text{ if }x>0\\0&\text{ otherwise.}\end{cases}\end{array}$$But that's not relevant here.

Answer 2

$\newcommand{nrm}[1]{\left\lVert{#1}\right\rVert}\newcommand{norm}{\nrm\bullet}$Let's call that norm $\norm^{C^1}_\infty$, whereas $\nrm{f}_\infty=\sup\limits_{x\in[a,b]}\lvert f(x)\rvert$; this way, $\nrm{f}_{\infty}^{C^1}=\nrm f_\infty+\nrm {f'}_\infty$.

The operator $$\frac d{dx}:\left(C^1[a,b],\norm^{C^1}_\infty\right)\to (C^0[a,b],\norm_\infty)$$ is, in point of fact, bounded: tautologically, $\nrm{\frac d{dx}}\le 1$.

What isn't bounded is the operator $$\frac d{dx}:\left(C^1[a,b],\norm_\infty\right)\to (C^0[a,b],\norm_\infty)$$ where $(C^1[a,b],\norm_\infty)$ is the subspace of $C^0[a,b]$ with the restricted norm. A glaring example of this is the sequence of functions $f_n(x)=\frac2\pi\arctan (nx)$ for the interval $[a,b]=[-1,1]$. $\nrm{f_n}_\infty\le1$, for all $n$, yet $\nrm{f'_n}_\infty=\sup\limits_{-1\le x\le 1}\frac2\pi\left\lvert\frac n{1+n^2x^2}\right\rvert=f'_n(0)=\frac{2n}\pi$.