Find norm of linear operator $Tf(t)=f(t)+\int_0^t(1-2s)f(s)ds$ Let $T:C[0,1]\to C[0,1]$ be linear operator defined as follows: $Tf(t)=f(t)+\int_0^t(1-2s)f(s)ds$. Find its norm where norm is is supremum norm.
I am not sure how to do it. It is easy to see that when we take $t\in[0,1/2)$, then we get best result if $f(t)=1$ or $f(t)=-1$. So we can guess than norm is $\leq1+1/4$. But I am not sure how to show that this guess is correct (if it actually is) so I would be thankful for hints how to go on.
Of course if $f$ is always equal to $1$, then we won't get anything better than $1+1/4$. So let's take $\epsilon>0$ and let's check $Tf(t+\epsilon)$ assuming that $f(t+\epsilon)<1$. Then $Tf(t+\epsilon)=f(t+\epsilon)+\int_0^{t+\epsilon}(1-2s)f(s)ds$. We want it to be $<1+1/4$. If $f(t+\epsilon)<0$ it is obvious. Otherwise $f(t+\epsilon)<1$ and $\int_0^{t+\epsilon}(1-2s)f(s)ds<1/4$. So we get what we want. We can use similar argument to show that $|Tf(t)|<1+1/4$ is bounded from both sides. Is it, rather unformal, reasoning on the right track?
 A: Formalizing some ideas by @user14717 and @SekstusEmpiryk from the comments:
\begin{align}
\|Tf\|_\infty &\le \|f\|_\infty + \max_{t \in [0,1]} \left|\int_0^t (1-2s)f(s)\,ds\right| \\
&\le \|f\|_\infty + \max_{t \in [0,1]} \int_0^t |1-2s|\underbrace{|f(s)|}_{\le \|f\|_\infty}\,ds \\
&\le \|f\|_\infty + \|f\|_\infty \int_0^1 |1-2s|\,ds
\end{align}
We have
$$\int_0^1 |1-2s|\,ds = \int_0^\frac12 (1-2s)\,ds + \int_{\frac12}^1 (2s-1)\,ds = \frac12$$
Therefore $$\|Tf\|_\infty \le \|f\|_\infty \left(1+\frac12\right) = \frac32 \|f\|_\infty$$
so $\|T\| \le \frac32$.
On the other hand, if you define $f_n \in C[0,1]$ as
$$f_n(t) = \begin{cases}
1, &\text{if } t \in \left[0,\frac12 - \frac1n\right]\\
-n\left(t-\frac12\right), &\text{if } t \in \left[\frac12 - \frac1n, \frac12+\frac1n\right]\\
-1, &\text{if } t \in \left[\frac12 - \frac1n, 1-\frac1n\right]\\
2n\left(x-1+\frac1n\right)-1, &\text{if } t \in \left[1 - \frac1n, 1\right]\\
\end{cases}$$
then we have $\|f_n\|_\infty = 1$ and 
\begin{align}
\|Tf\|_\infty &\ge Tf(1) \\
&= 1 + \underbrace{\int_0^{\frac12-\frac1n} (1-2s)\,ds}_{=\frac14 - \frac1{n^2}} + \underbrace{\int_{\frac12-\frac1n}^{\frac12+\frac1n} (1-2s)f(s)\,ds}_{\ge 0} + \underbrace{\int_{\frac12+\frac1n}^{1-\frac1n} (2s-1)\,ds}_{=\frac14 - \frac1{n}} + \underbrace{\int_{1-\frac1n}^1 (1-2s)f(s)\,ds}_{\ge -\frac1n}\\
&\ge 1+\frac14 - \frac1{n^2} + \frac14 - \frac1{n} -\frac1n\\
&= \frac32 -\frac{2}n - \frac1{n^2}
\end{align}
So we conclude
$$\|T\| \ge \frac{\|Tf_n\|_\infty}{\|f_n\|_\infty} \ge \frac32 -\frac{2}n - \frac1{n^2} \xrightarrow{n\to\infty} 1$$
Hence $\|T\| = \frac32$.
A: To give a bound note:
$$||T(f)(t)|| \leq ||T(f)(t)||_{\infty} \leq ||f(t)||_{\infty} +  || \int\limits_0^t (1-2s)f(s) ds||_{\infty}.$$
The integral we can write out as: 
$$|| \int\limits_0^t (1-2s)f(s) ds||_{\infty} \leq  \int\limits_0^t ||(1-2s)f(s)||_{\infty}ds \leq t ||(1-2t)||_{\infty} ||f(t)||_{\infty} =||f(t)||_{\infty}.$$
So, we can conclude: $||T||\leq 2$ 
