Some time ago my teacher showed the solution of this exercise. Today I reviewed it, and I think he might be wrong at the last part, c.)
Exercise: Let $a > 0$ and let $g \in C[0,a]$ be a non-negative function. Consider the operator $T: C[0,a] \rightarrow C[0,a]$, defined by
$(Tf)(x) = \int_0^x g(t)f(t)dt\;.$
a.) Prove that the operator $T$ is linear.
Solution: $\begin{align*} T(f+h) &= \displaystyle \int_0^x g(t)(f+h)(t)dt = \int_0^x g(t)f(t) + g(t)h(t)dt\\ &= \int_0^x g(t)f(t)dt + \int_0^x g(t)h(t)dt = T(f) + T(h)\;;\\ T(\alpha f) &= \displaystyle \int_0^x \alpha \; g(t)f(t)dt = \alpha \int_0^x g(t)f(t)dt = \alpha \; T(f)\;. \end{align*}$
b.) Prove that the operator $T$ is continuous.
Solution: Show that it is bounded (which implies it is continuous), i.e. some constant $c$ should exist such that $\|Tf\|_\infty \le c\; \|f\|_\infty$:
$\|Tf\|_\infty \le (\text{interval length})\|g\|_\infty\|f\|_\infty = a \; \|g\|_\infty\|f\|_\infty\;;$ in this case $c = a \; \|g\|_\infty$.
c.) Define the norm of the operator $T$. The idea is to show that the inequality from b.) could change to an equality if you pick the right function for $f(t)$. He said, choose $f(t) = \dfrac{\|g\|_\infty}{g(t)}$. That way, $Tf = x\; \|g\|_\infty$; the sup-norm of this is of course $a \; \|g\|_\infty$.
Definition of the norm of an operator $T$: $\|T\| = \sup_{x\in \mathcal{D}(T)} \frac{\|Tf\|_\infty}{\|f\|_\infty}\;.$ We saw that $\sup \|Tf\|_\infty = a \; \|g\|_\infty$, so $\|f\|_\infty$ should be $1$, right? But this isn't the case:
$\|f\|_\infty = \left\| \dfrac{\|g\|_\infty}{g(t)} \right\|_\infty \neq 1\;.$ However, $g(t) / \|g\|$ would be.
So, what's going wrong here? The sup-norm of a function in $C[a,b]$ is just the maximum value of that function on the given interval, right. I'm not entirely sure about the definition of $\|T\|$; the sup-part seems redundant?
[Edit] I just added a comment, since $\|f(t)g(t)\|_\infty = \|g(t)\|_\infty$ this would result in the expected value for the norm. However, in order to calculate the norm of $T$ you should to be able to divide $\|Tf\|$ by $\|f\|$. However, since $\|f\|$ can go to infinity, I don't see how I should this. And once more, I don't understand why there should be a $\sup$ condition for the $\|T\|$ since it is already a quotient of norms, so the $\sup$ condition could be left out in my opinion?