Computation of linear operator norm. $||T||1=sup||f||≤1,||f||∈C(J)||Tx||$

Question

Let $J=[0,a], a>0$ with $C(J)$ with norms $||f||_{1}=\max_{x\in J}|f(x)|e^{-ax}$

$T : C(J) \longmapsto C(J) $ is defined by $Tf(x)=\int_0^x \! tf(t) \, \mathrm{d}t.$

Then, compute $||T||_{1}$.

So, norm of linear operator is given by

$||T||_{1}=\sup_{||f|| \leq 1,||f|| \in C(J)}||Tx|| $

I think it follows $||Tx||=\max_{x \in J}|Tf(x)|e^{-ax}$

Thus, it seems like that $||T||_{1}=\sup_{||f|| \leq 1,||f|| \in C(J)}(\max_{x \in J}|Tf(x)|e^{-ax})$

I am not even sure whether I am on right track. If I am, is there any nice way to deal with $\sup$ and $\max$ ?

Answer 1

You have to find the supremum of $\max e^{-ax} |\int_0^{x} tf(t)dt|$ subject to the condition $|f(x)| \leq e^{ax}$. Clearly the maximum value is obtained by taking $f(x) = e^{ax}$. To maximize $ e^{-ax} \int_0^{x} te^{at}dt$ you can compute the integral explicitly (by integrating by parts). This gives $\frac 1 {a^{2}} e^{-ax} [1+a-e^{ax}-axe^{ax}]=\frac 1 {a^{2}} [(1+a)e^{-ax}-1-ax]$. This function is decreasing and its maximum value is attained at $0$. The norm of $T$ is therefore equal to $a$.

Answer 2

I don't have the rep points to comment, so I'll use an answer box. One thing that might help, notice that the set of functions such that $||f||_1 = 1$ under that max norm are all $f(x)$ such that $|f(x)| \le e^{ax}$ [ETA: and $|f(x)|$ must equal $e^{ax}$ at at least one point]. You can play around with that and see when it gives you the largest value in $\max_{x \in J} |Tf(x)|e^{-ax}$ (I suspect at least it's when $f(t)=e^{at}$; you'd still have to find $x$).