I have question about solving the norm of a linear operator. I have my own solution that is different from the formal solution. The formal solution is a little difficult compared with my solution. I want to if I have made a mistake, or why does the formal solution take my method?
The problem can be stated as:
Let T be an operator from $L[a,b]$ to $L[a,b]$, $f\in L[a,b]$, define $$(Tf)(x)=\int_{a}^{x}f(t)dt.$$ Proof that the norm of $T$ is $b-a$.
My solution is as following
$\|Tf\|=\int_{a}^{b}|\int_{a}^{x}f(t)dt|dx\le\int_{a}^{b}\int_{a}^{x}|f(t)|dtdx\le\int_{a}^{b}\int_{a}^{b}|f(t)|dtdx=(b-a)\|f\|$. So $\|T\|\le b-a.$ Then we select a function $f_0(t)=\frac{1}{b-a},$ we can achieve $\|T\|\ge b-a$. Therefore, I proof it. So where is my mistake? I'm puzzled. The formal solution to the problem is not like this.
If $f_0(t)=\frac1{b-a}$, then$$\lVert f_0\rVert=\int_a^b\frac1{b-a}\,\mathrm dx=1.$$But$$T(f_0)(x)=\int_a^x\frac1{b-a}\,\mathrm dt=\frac{x-a}{b-a}$$and therefore$$\bigl\lVert T(f_0)\bigr\rVert=\int_a^b\frac{x-a}{b-a}\,\mathrm dx=\frac{b-a}2,$$which is smaller than $b-a$.