Proof verification: continuity of $T$ on $\mathcal{C}([0,1],\mathbb{R})$

Question

Given $$T: \mathcal{C}([0,1],\mathbb{R}) \mapsto \mathcal{C}([0,1],\mathbb{R}) \qquad \qquad f \mapsto \Bigg(t \mapsto \sin\bigg(\int_{0}^{t} f(s) \ ds\bigg)\Bigg)$$

where $\mathcal{C}([0,1],\mathbb{R})$ is equipped by Sup Norm. Investigate the continuity of $T$

My work:

$T$ is discontinious on $\mathcal{C}([0,1],\mathbb{R})$

I will show this by proving that there exists an $\epsilon > 0$ such that for all $\delta > 0$ and all $f, g \in \mathcal{C}([0,1],\mathbb{R})$ : $d(f,g) < \delta$ but $d(T(f),T(g)) > \epsilon$

Let $\epsilon := 1$ and $\delta > 0$

Assume $f, g \in \mathcal{C}([0,1],\mathbb{R})$ such that $d(f,g) = \text{sup}_{x \in [0,1]}{\big| \ f(x)-g(x)\big|} < \delta$

It follows that $d(T(f),T(g)) = \text{sup}_{x, s \in [0,1]}\Big\{\Big|\sin\int_{0}^{x} f(s) \ ds - \sin\int_{0}^{\tilde{x}} g(s) \ ds\Big|\Big\} \leq \text{sup}\Big\{ \Big|\sin\int_{0}^{x} f(s) \ ds\Big| + \Big|\sin\int_{0}^{\tilde{x}} g(s) \ ds\Big| \Big\} \\ \leq \text{sup}\Big\{ \Big|\sin\int_{0}^{x} f(s) \ ds\Big|\Big\} + \text{sup}\Big\{\Big|\sin\int_{0}^{\tilde{x}} g(s) \ ds\Big| \Big\} \leq 2 \nless \epsilon $

Is my proof correct? Thanks.

Answer 1

Not really, it should be $$d(Tf,Tg) = \|Tf-Tg\|_\infty = \sup_{x\in[0,1]}\left|\sin\int_{0}^{x} f(s) \ ds - \sin\int_{0}^{{x}} g(s) \ ds\right| \le 2$$ and it doesn't really show anything because $\|Tf-Tg\|_\infty \le 2$ doesn't imply that it is $\|Tf-Tg\|_\infty> \varepsilon$.

Your operator is in fact continuous. Indeed, it is a composition of two maps $A,B : C[0,1] \to C[0,1]$ where $$Af = \int_0^\cdot f(s)\,ds, \quad Bf = \sin\circ\, f$$ $A$ is a bounded linear map $$\|Af\|_\infty = \sup_{x\in[0,1]}\left|\int_0^x f(s)\,ds \right| \le \int_0^1 |f(s)|\,ds \le \|f\|_\infty$$ and hence continuous.

To show that $B$ is (uniformly) continuous, let $\varepsilon > 0$. Since $\sin$ is uniformly continuous on $[0,1]$, pick $\delta> 0$ such that $\|x-y\| < \delta \implies \left|\sin x - \sin y\right| < \varepsilon$.

For any $f,g \in C[0,1]$ with $\|f-g\|_\infty < \delta$ we have $|f(x) - g(x)| < \delta$ for any $x\in[0,1]$ so $$\|Bf-Bg\|_\infty = \sup_{x\in[0,1]}\left|\sin f(x) - \sin g(x)\right| \le \varepsilon$$

Hence $B$ is continuous. Now, $T = B\circ A$ so it is continuous.

Answer 2

No, your proof isn’t correct.

To prove $T$ discontinuity, you should find $f \in V= \mathcal C([0,1], \mathbb R)$, $\epsilon >0$ and a sequence $(f_n) \in V$ such that $d(f_n, f) \to 0$ and $d(T(f_n),T(f)) >\epsilon$ for all $n \in \mathbb N$. This would contradict the $\epsilon$-$\delta $ definition of continuity.

In fact $T$ is continuous as you have for $f,g \in V$ and $t \in [0,1]$

$$\begin{aligned} \vert T(f)(t) - T(g)(t) \vert &=\left\vert \sin \left(\int_{0}^{t} f(s) \ ds \right) - \sin \left(\int_{0}^{t} g(s) \ ds \right)\right\vert\\ &\le \left\vert \left(\int_{0}^{t} f(s) \ ds \right) - \left(\int_{0}^{t} g(s) \ ds \right)\right\vert \\ &=\left\vert \int_0^t \left(f(s)-g(s) \right) \ ds\right\vert\\ &\le \int_0^t \left\vert f(s)-g(s) \right\vert \ ds\\ &\le \int_0^1 \left\vert f(s)-g(s) \right\vert \ ds \le d(f,g) \end{aligned}$$

As this inequality is valid for all $t\in [0,1]$,

we can conclude that $d(T(f),T(g)) \le d(f,g)$ proving that $T$ is a short map hence continuous.