This is the question that was given to me:
I was able to prove that the operator is bounded: it is bounded by the infinity norm of $f$.
Since it is bounded, it has a finite norm, $$\|T\|= \sup\limits_{f\in C[0,1]} \frac{\|Tf\|}{\|f\|}. $$
I noticed that for $f=1$ this is equal to $1/2$. I think that I need to find an upper bound, and show that it is attained, probably with $f=1$, but I am stuck as to what this bound might be. Any help is appreciated.
Best Answer
Let $\|\cdot\|_\infty$ denote the sup norm. For all $x\in[0,1]$, $$\begin{align*}|Tf(x)|&=\left|\int_0^x (x-t)f(t)\,dt\right|\\ &\leq \int_0^x|(x-t)f(t)|\,dt\\ &\leq \int_0^x(x-t)\|f\|_\infty\,dt\\ &=\frac12 x^2\|f\|_\infty\\ &\leq \frac12\|f\|_\infty \end{align*},$$ hence $\|Tf\|_\infty\leq \frac12\|f\|_\infty$. This shows that $\|T\|\leq \frac12$, and your example with $f=1$ shows $\|T\|\geq \frac12$, so the norm must equal $\frac12$.