I'm trying to solve an exercise in Brezis' Functional Analysis
Let $E:=C([0, 1])$ be the Banach space with the supremum norm $\|\cdot\|_\infty$. We define a bounded linear operator $T:E\to E$ by $$ Tu(t) := \int_0^t u(s) \, \mathrm d s \quad \forall u \in E, t \in [0, 1]. $$ Let $B$ be the closed unit ball of $E$. Then $T(B)$ is not closed.
There are possibly subtle mistakes that I could not recognize in below attempt. Could you have a check on it? Thank you so much for your help!
Let $u_n (s) := |s-\frac{1}{2}|^{1/n}$ and $v (s):= |s-\frac{1}{2}|$ for $n \ge 1$ and $s \in [0, 1]$. Then $u_n \in B, v \in E$ and $Tu_n (s) =|s-\frac{1}{2}|^{1/n+1}$. Then $\|Tu_n-v\|_\infty \to0$. Clearly, $v$ is not differentiable at $1/2$. On the other hand, $T(B) \subset C^1([0, 1])$. This completes the proof.
Update 1 I made explicit computations as follows. Then I have realized that $v$ is smooth!!!
Let $u_n (s) := |s-\frac{1}{2}|^{1/n}$ and $v_n := Tu_n$. For $t \le 1/2$, we have $$ \begin{align} v_n (t) &= \int_0^t \left ( \frac{1}{2} -s \right )^{1/n} \, \mathrm d s \\ &= -\int_{\frac{1}{2}}^{\frac{1}{2}-t} s^{1/n} \, \mathrm d s \\ &= \int^{\frac{1}{2}}_{\frac{1}{2}-t} s^{1/n} \, \mathrm d s \\ &= \frac{1}{\frac{1}{n}+1} s^{\frac{1}{n}+1} \bigg |^{\frac{1}{2}}_{\frac{1}{2}-t} \\ &= \frac{n}{n+1} \left ( \frac{1}{2^{\frac{1}{n}+1}} - \left (\frac{1}{2} -t \right )^{\frac{1}{n}+1} \right ). \end{align} $$
For $t > 1/2$, we have $$ \begin{align} v_n (t) &= \int_0^{1/2} \left ( \frac{1}{2} -s \right )^{1/n} \, \mathrm d s + \int_{1/2}^t \left ( s-\frac{1}{2} \right )^{1/n} \, \mathrm d s \\ &= \frac{n}{n+1}\frac{1}{2^{\frac{1}{n}+1}} + \int_{1/2}^t \left ( s-\frac{1}{2} \right )^{1/n} \, \mathrm d s \\ &= \frac{n}{n+1}\frac{1}{2^{\frac{1}{n}+1}} + \frac{1}{\frac{1}{n}+1} \left ( s-\frac{1}{2} \right )^{\frac{1}{n}+1} \bigg |_{\frac{1}{2}}^t \\ &= \frac{n}{n+1} \left (\frac{1}{2^{\frac{1}{n}+1}} + \left ( t-\frac{1}{2} \right )^{\frac{1}{n}+1} \right ). \end{align} $$
Let $v(t) := t$ for $t \in [0, 1]$. Then $\|v_n -v\|_\infty \to 0$.
My computation in Update is actually correct, so $u_n (t) = |t-\frac{1}{2}|^{1/n}$ will not work for the proof. As @Ryszard suggested in a comment, we let $v_n(t):=1/2^{1/n+1}-|t-1/2|^{1/n+1}$ and $v(t) := 1/2-|t-1/2|$ for $t\in [0, 1]$. Then $\|v_n -v\|_\infty \to 0$. Also, $v \in E$ is not continuously differentiable. Let $u_n (t) := v'_n (t)$. By FTC, $Tu_n (t) =\int_0^t u_n(s) \, \mathrm d s =v_n(t)-v_n(0) =v_n(t)$. Indeed, $$ u_n (t) = - \frac{n+1}{n} \operatorname{sgn} \left (t-\frac{1}{2} \right ) \left |t- \frac{1}{2} \right |^{1/n} \quad \forall t \in [0, 1]. $$