Showing that the derivative function is not continuous w.r.t. the sup norm

Question

Let $\mathcal{C}([-1, 1]) = \{f: [-1, 1] \to \mathbb{R}\mid f \text{ is continuous}\}$, $\mathcal{C}^{1}([-1, 1]) = \{f: [-1, 1] \to \mathbb{R}\mid f' \text{ is continuous}\}$, where both sets are equipped with the supremum norm $||\cdot||_\infty$. Let $h: \mathcal{C}^{1}([-1, 1]): \to \mathcal{C}([-1, 1]), [h(f)](t) = f'(t), t \in [-1, 1]$ be the derivative mapping from the set of continuously differentiable functions to continuous functions. I want to show that $h$ is not a continuous mapping w.r.t. the supremum norm, but I have not yet managed to "hack" the intuition for functional analysis. Specifically, what we want to demonstrate is that $\exists \epsilon > 0: \forall \delta > 0: \forall f, g \in \mathcal{C}^{1}([-1 ,1]): d_{\mathcal{C}^{1}}(f, g) < \delta\land d_{\mathcal{C}}(h(f), h(g)) \geq \epsilon$.

Where I am currently stuck is that before inspecting the derivative mapping, let $f(t) = \cos(t)$ and $h(t) = 0$. Both are evidently continuous, but $||f - h||_\infty = \sup_{t \in [-1, 1]}\left|\cos(t) - 0\right| = 1$. So to my understanding, even a redefined $[h(f)](t) = f(t)$ should not be continous, as for no $\delta$ is $1 < 0.5$. Or am I mistaken? In the case that I am correct, does this same claim show that the original (derivative mapping) $h$ is not continuous?

Answer 1

Let $f_n(t): \frac{1}{n} \sin (nt).$ Then $||f_n||_{\infty} \le \frac{1}{n}$ for all $n$. Hence

$$ ||f_n -0||_{\infty} \to 0.$$

But we have $(h(f_n))(t)= \cos (nt)$ and therefore

$(h(f_n)))$ does not converge to $(h(0))$ in the norm $||\cdot||_\infty$

Answer 2

Continuity of $h$ means given $f$ and $\epsilon >0$ there exist $\delta >0$ such that $\|g-f\|_{\infty} <\delta$ implies $\|h(f)-h(g)\|_{\infty} <\epsilon$. In your example $\|h-f\|_{\infty} <\delta$ is not satisfied for any $\delta \in (0,1)$ so the example says nothing about continuity of $h$. [BTW, you should avoid using $h$ with two diferent meanings].

To prove that $h$ is not continuous consider the funsions $f_n(x)=\frac {x^{n}} n$. These converge in the norm to the $0$ function but their derivatives do not converge in the norm to the $0$ function since $\sup\{|x^{n-1}-0|: 0 \leq x \leq 1\}=1$ for all $n$.