Dtyjlui

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

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|sinint_{0}^{x} f(s) ds - sinint_{0}^{tilde{x}} g(s) dsBig|Big} leq text{sup}Big{ Big|sinint_{0}^{x} f(s) dsBig| + Big|sinint_{0}^{tilde{x}} g(s) dsBig| Big} \ leq text{sup}Big{ Big|sinint_{0}^{x} f(s) dsBig|Big} + text{sup}Big{Big|sinint_{0}^{tilde{x}} g(s) dsBig| Big} leq 2 nless epsilon $

Is my proof correct? Thanks.

Not really, it should be $$d(Tf,Tg) = |Tf-Tg|_infty = sup_{xin[0,1]}left|sinint_{0}^{x} f(s) ds - sinint_{0}^{{x}} g(s) dsright| le 2$$
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
$$Af = int_0^cdot f(s),ds, quad Bf = sincirc, f$$
$A$ is a bounded linear map
$$|Af|_infty = sup_{xin[0,1]}left|int_0^x f(s),ds right| le int_0^1 |f(s)|,ds le |f|_infty$$
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 $|x-y| < delta implies left|sin x - sin yright| < varepsilon$.

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

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

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 $t in [0,1]$

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

As this inequality is valid for all $tin [0,1]$,

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