Continuity of the differential operator from $(C^1[0,1],d_\infty)$ to $(C[0,1],d_1)$

Question

Let $C^1[0,1]$ denote the set of continuously differentiable functions on the unit interval , let $C^0[0,1]$ denote the set of continuous functions on the unit interval. Let $d_\infty$ and $d_1$ denote the metrics given by:

$d_\infty(f,g)=\sup_{x \in [0,1]}|f(x)-g(x)|$,

$d_1(f,g)=\int_0^1 |f(x)-g(x)|dx$.

Is the differential operator $D:(C^1[0,1],d_\infty)\mapsto(C^0[0,1],d_1):f\mapsto f'$ continuous? Would the answer change if we restrict to the set of continuously differentiable convex functions on [0,1] (endowed with $d_\infty$) and endow $C^0[0,1]$ with $d_\infty$?

Answer 1

The answer is NO.

Let $n\in \mathbb{N}$ be large. Consider the points $$ E_n = \left\{0, \frac{1}{n}, \frac{2}{n},..., \frac{n-1}{n}, 1\right\}, $$ and define a function $F_n$ on $[0,1]$ to be $F_n = 0$ on $E_n$ and if $\Delta_k = [\frac{k}{n}, \frac{k+1}{n}]$ where $0\leq k < n$, then define $F_n$ to be $1/n$ at the midpoint of $\Delta_k$ and linear on the rest of the interval $\Delta_k$. So effectively the graph of $F_n$ looks like a small triangle with height $1/n$ and having equal edges.

Clearly $d_\infty( F_n, 0 ) = 1/n \to 0$, and $F_n$ is differential at all points of $[0,1]$ except points of $E_n$ and the midpoints of the triangles $\Delta_k$. We thus have $|F_n'(x)| = 1$ for all $x \in [0,1] \setminus ( E_n \cup (E_n + 1/2n) ) $, and hence $d_1(F_n', 0 ) = 1$, thus the differentiation operator is not continuous.

Of course, $F_n$ is not continuously differentiable, but it is straightforward to "smoothen" $F_n$ at the points of $E_n$ and the midpoints of the triangles $\Delta_k$ to get a $C^1$ function violating the continuity of the operator $D$.

Answer 2

To complete the discussion, let's now consider the set $\mathscr{C}^1[0,1]$ of convex continuously differentiable functions on [0,1]. Clearly, for an arbitrarily small $\epsilon_0>0$, we can consider each element $f \in \mathscr{C}^1[0,1]$ as the restriction on $[0,1]$ of a finite element of $ \mathscr{C}^1(-\epsilon_0,1+\epsilon_0)$. By Theorem 25.7 in the book Convex Analysis by Rockafellar (1972), we have that for any sequence $(f_n)$ and function $f$ which are finite elements of $\mathscr{C}^1(-\epsilon_0,1+\epsilon_0)$, $f_n\to f$ pointwise on $(-\epsilon_0,1+\epsilon_0)$ entails that $d_\infty(Df_n,Df)\to0$ - recall that the supremum is evaluated on $[0,1]$. Consequently, the differential operator is sequentially continuous as a map from $(\mathscr{C}[0,1], d_\infty)$ to $(C^0[0,1], d_\infty)$. Finally, since sequential continuity is equivalent to continuity in metric spaces, we can conclude that the map $D:(\mathscr{C}^1[0,1], d_\infty)\to(C^0[0,1], d_\infty)$ is continuous.