Csdrhrt

Let $mathcal{C}$ be the space of continuous functions on $[0,1]$ equipped with the norm $|f|=int^{1}_{0}|f(t)|dt$. Define a linear map $mathcal{T}:mathcal{C}rightarrow mathcal{C}$ by
$$ mathcal{T}f(x)=int^{x}_{0}f(t)dt. $$
Show that $mathcal{T}$ is well-defined and bounded and determine the value of $|mathcal{T}|_{text{op}}$.

I proved the first two parts myself, but I am having trouble with determining the value of $|mathcal{T}|_{text{op}}$. I was able to show that it is bounded by $1$ though. Observe that

$$|mathcal{T}f|=int^{1}_{0}left|int^{x}_{0}f(t)dtright|dxleq int^{1}_{0}|f|dx = |f| $$
Therefore,

$$ |mathcal{T}|_{text{op}} = underset{|f|=1}{sup}frac{|mathcal{T}f|}{|f|}leq underset{|f|=1}{sup}frac{|f|}{|f|}=1 $$

I tried seeing if I could then construct a function where the operator equals $1$, but I've had no success. Anybody have any solutions or hints? Any help is appreciated.

For $nin Bbb N:$ Let $K_n=frac {1}{n+1}+frac {1}{(n+1)^2}.$ Let $f_n(x)=n+1$ for $xin [0,frac {1}{n+1}].$ Let $f_n(x)=0$ for $xin [K_n,1].$ Let $f_n(x)$ be linear for $xin [frac {1}{n+1},K_n].$

We have $|f_n|=1+frac {1}{2(n+1)}.$

For $xin [frac {1}{n+1},1]$ we have $(Tf_n)(x)geq (Tf_n)(frac {1}{n+1})=1.$ $$text {So }quad |Tf_n|geq int_{1/(n+1)}^1 (Tf_n)(x)dxgeq int_{1/(n+1)}^1 1cdot dx=$$ $$=1-frac {1}{n+1}.$$

$$text {So} quad frac {|Tf_n|}{|f_n|}geq frac {1-frac {1}{n+1}}{ 1+frac {1}{2(n+1) }}.$$

It might be hard (or impossible) to find a function for which $ ||mathcal{T}f||$ is exactly equal to $||f||$, but you only need to find a sequence of functions $f_n$, such that $$frac{||mathcal{T}f_n||}{||f_n||} to 1quad text{as } n to infty.$$

Note that you do not need the limit of the $f_n$ to be a continuous function!

You should try to construct one such sequence using elementary functions. Try to write down a few examples with some free parameters, and see if you can cook up such a sequence!