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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?

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    Perhaps it has something to do with the step $\|Tf\|_\infty \le a \; \|f(t)g(t)\|_\infty \le a \; \|f(t)\|_\infty \|g(t)\|_\infty$...? Because $\|f(t)g(t)\|_\infty = \|g\|_\infty$, but $\|f(t)\|_\infty\|g(t)\|_\infty$ can be infinite, right.2011-10-31

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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: $\|Tf_1\|_\infty = \frac{1}{\|f\|_\infty}\|Tf\|_\infty = \alpha \|g\|_\infty = \|T\|$

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    @Ailurus $\|Tf\|_\infty = \alpha \|g\|_\infty$ does not imply $\|f\|_\infty = 1$, and as you have pointed out it is actually not true in this case. There is nothing in the definitions excluding the possibility that a linear operator could achieve its norm for a non-unit vector. It is only asserted that it does achieves it norm for some unit vector also, as above.2011-11-01