Bounded Linear Operator from $C_0([0,1])$ to $C([0,1])$ 
Define
$$
C_0([0,1]) := \left\{f\in C([0,1]) : \int_0^1f(t)\, \mathrm dt=0\right\}.
$$
Show that $T : C_0([0,1]) \to C([0,1])$, given by
$$
(Tf)(x) := \int_0^x(t-x) f(t) \, \mathrm dt, 
$$
defines a bounded linear operator.

I proved $\Vert T \Vert \leq \frac 1 2$, but I could not find some $f \in C_0([0,1])$ such that $\Vert f \Vert = 1$ and $\Vert T f \Vert = \frac 1 2$.
Here is how to prove $\|T\|\leq \frac12$.
Note that
$$\|Tf\|=\sup_{x\in[0,1]}\,\left|\int_0^x\,(t-x)\,f(t)\,\text{d}t\right|\leq \sup_{x\in [0,1]}\,\int_0^x\,|t-x|\,\big|f(t)\big|\,\text{d}t\,.$$
Since $\big|f(t)\big|\leq \|f\|$ for all $t\in[0,1]$, we get
$$\|Tf\|\leq \sup_{x\in[0,1]}\,\int_0^x\,|t-x|\,\|f\|\,\text{d}t\leq \sup_{x\in[0,1]}\,\|f\|\,\int_0^x\,(x-t)\,\text{d}t\,.$$
Therefore,
$$\|Tf\|\leq \|f\|\,\sup_{x\in[0,1]}\,\int_0^x\,(x-t)\,\text{d}t\leq \|f\|\,\sup_{x\in [0,1]}\,\int_0^x\,(1-t)\,\text{d}t\,.$$
So,
$$\|Tf\|\leq \|f\|\,\int_0^1\,(1-t)\,\text{d}t=\frac{1}{2}\,\|f\|\,.$$
 A: Let $f\in \mathcal{C}_0\big([0,1]\big)$.  Fix $x\in[0,1]$.  Note that
$$-Tf(x)=\int_0^x\,(x-t)\,f(t)\,\text{d}t=\int_0^x\,\int_t^x\,f(t)\,\text{d}s\,\text{d}t=\int_0^x\,\int_0^s\,f(t)\,\text{d}t\,\text{d}s\,.$$
Write
$$If(x):=\int_0^x\,f(t)\,\text{d}t\,.$$
Then,
$$-Tf(x)=\int_0^x\,If(s)\,\text{d}s\,.$$
Observe that
$$\big|If(x)\big|\leq \int_0^x\,\big|f(t)|\,\text{d}t\leq \|f\|_\infty\,x$$
and
$$\big|If(x)\big|=\big|If(1)-If(x)\big|=\left|\int_x^1\,f(t)\,\text{d}t\right|\leq \|f\|_\infty\,(1-x)\,.$$
Thus,
$$\big|If(x)\big|\leq \|f\|_\infty\,\min\{x,1-x\}\,.$$
Ergo,
$$\big|Tf(x)\big|\leq \int_0^x\,\big|If(s)\big|\,\text{d}s\leq \|f\|_\infty\,\int_0^x\,\min\{s,1-s\}\,\text{d}s\,.$$
This implies
$$\big\|Tf\big\|_\infty\leq \|f\|_\infty\,\int_0^1\,\min\{s,1-s\}\,\text{d}s=\frac{1}{4}\,\|f\|_\infty\,.\tag{*}$$
Therefore, $\|T\|_\text{op}\leq \dfrac14$.
Note that (*) is an equality if and only if $f\equiv 0$, since the equality case occurs only when there exists $\epsilon$ such that $|\epsilon|=1$ and $If(x)=\epsilon\,\|f\|_\infty\,\max\{x,1-x\}$ for all $x\in[0,1]$.  This condition, along with the fact that $f$ is continuous, implies that $f\equiv 0$ is the only possible choice.  However, we can find a sequence $\left(f_n\right)_{n=1}^\infty$ of functions $f_n\in\mathcal{C}_0\big([0,1]\big)$ satisfying $\|f_n\|_\infty=1$ and $\lim\limits_{n\to\infty}\,\|Tf_n\|_\infty=\dfrac14$.  For each $n=1,2,3,\ldots$, take
$$f_n(x)=\left\{
\begin{array}{ll}
-1&\text{if }0\leq x\leq \dfrac{1}{2}-\dfrac1{2n}\,,\\
2n\left(x-\dfrac12\right)&\text{if }\dfrac{1}{2}-\dfrac{1}{2n}<x<\dfrac12+\dfrac1{2n}\,,\\
+1&\text{if }\dfrac12+\dfrac1{2n}\leq x\leq 1\,.
\end{array}
\right.$$
Observe that
$$\|Tf_n\|_\infty=Tf_n(1)=\dfrac14-\dfrac1{12n^2}$$
for every positive integer $n$.
