Norm of the linear bounded operator $T$ defined by $(Tf)(x) = \int_0^x g(t)f(t)dt$ 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?
 A: You are confusing the two suprema mentioned in the definition. The norm of an element $f \in C(0, \alpha)$ is given by $$\|f\|_\infty = \sup_{0 \le x \le \alpha} |f(x)|$$
However, the norm of a linear operator $T:  C(0, \alpha) \to  C(0, \alpha)$ is given by
$$\|T\| = \sup_{f \in C(0, \alpha)} \frac{\|Tf\|_\infty}{\|f\|_\infty}$$
To clarify, in what you wrote,
$$\|T\| = \sup_{f \in D(T)} \frac{\|Tf\|_\infty}{\|f\|_\infty}$$
$D(T)$ stands for the domain of $T$, which in this case is all of $C(0,\alpha)$.
So in the definitions the suprema are taken of different functions over different domains. There is no redundancy.
So what's going on is the following:
we have
$$\|Tf\|_\infty = \sup_{0 \le x \le \alpha}|\int_0^x g(t)f(t) dt| $$
$$ \le \int_0^\alpha \sup_{0 \le t \le \alpha} |f(t)||g(t)| dt$$
$$ \le \alpha \|f\|_\infty \|g\|_\infty$$
Hence 
$$\|T\| = \sup_{f \in C(0, \alpha)} \frac{\|Tf\|_\infty}{\|f\|_\infty} \le \sup_{f \in C(0, \alpha)} \frac{\alpha \|f\|_\infty \|g\|_\infty}{\|f\|_\infty} = \alpha \|g\|_\infty$$
So to show that this upper bound can also be obtained, i.e., it is the actual supremum, we must exhibit an element $f \in C(0, \alpha)$ such that
$$ \frac{\|Tf\|_\infty}{\|f\|_\infty} = \alpha \|g\|_\infty$$
Take (assuming $g(t) > 0$ like in the comments) $f = \frac{\|g\|_\infty}{g}$. Then
$$Tf = \int_0^x g(t) \frac{\|g\|_\infty}{g(t)} dt = \int_0^x \|g\|_\infty dt = x\|g\|_\infty$$
So $$\|Tf\|_\infty = \sup_{0 \le x \le \alpha} \|g\|_\infty x = \alpha \|g\|_\infty$$
I hope those details made things more clear. Let me know in the comments if this was not what you were looking for.
EDIT: I'm guessing from what you wrote in the last part of your question that you also saw the following definition of operator norm:
$$\|T\| = \sup_{\|x\| = 1} \|Tx\|$$
It is simple to prove that this is equivalent to the definition given above. In fact, we just showed that for $f = \frac{\|g\|_\infty}{g}$ we get
$$\|Tf\|_\infty  = \alpha \|g\|_\infty \|f\|_\infty = \|T\|\|f\|_\infty$$
so the unit vector $f_1 = \frac{f}{\|f\|_\infty}$ achieves the supremum over the unit sphere:
$$\|Tf_1\|_\infty = \frac{1}{\|f\|_\infty}\|Tf\|_\infty = \alpha \|g\|_\infty = \|T\|$$
