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