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?

  • $\begingroup$ the function $g$ is suppose to be only continuous and non-negative ? If so $f$ you propose it not always defined in whole $[a,b]$, if $g$ vanish in this interval. $\endgroup$ – Leandro Oct 31 '11 at 0:16
  • $\begingroup$ @Leandro, yep that's a little strange indeed. My teacher also mentioned that it should be better if $g(t)$ would be strictly positive. Brian, thanks for fixing the long $\LaTeX$ equation. $\endgroup$ – Ailurus Oct 31 '11 at 0:31
  • $\begingroup$ 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. $\endgroup$ – Ailurus Oct 31 '11 at 13:43

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

  • $\begingroup$ This was the solution given to the OP by his/her teacher. It doesn't quite work since, as noted by the OP, $||f||\neq 1$ necessarily. $\endgroup$ – Kevin Ventullo Oct 31 '11 at 19:06
  • $\begingroup$ @Kevin yeah, but the point is you don't need $\|f\|=1$ in the first definition of operator norm. And if you choose to work with the second definition, $f/\|f\|$ does the trick. $\endgroup$ – user12014 Oct 31 '11 at 19:09
  • $\begingroup$ I'm still a little confused about the part where you say we have to show that $\dfrac{\|Tf\|_\infty}{\|f\|_\infty} = \alpha \; \|g\|_\infty$, but two equations later $\|Tf\|_\infty = \alpha \|g\|_\infty$, implying $\|f\|_\infty$ should equal $1$, while this isn't the case (unless you divide f by its sup norm). The rest is clear, thanks for explaining. $\endgroup$ – Ailurus Nov 1 '11 at 11:29
  • $\begingroup$ @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. $\endgroup$ – user12014 Nov 1 '11 at 23:33

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