Value of operator norm when $\mathcal{T}f(x)=\int^{x}_{0} f(t)dt$ 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
$$\|\mathcal{T}f\|=\int^{1}_{0}\left|\int^{x}_{0}f(t)dt\right|dx\leq \int^{1}_{0}\|f\|dx = \|f\| $$
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.
 A: For $n\in \Bbb N:$ Let $K_n=\frac {1}{n+1}+\frac {1}{(n+1)^2}.$ Let $f_n(x)=n+1$ for $x\in [0,\frac {1}{n+1}].$ Let $f_n(x)=0$ for $x\in [K_n,1].$ Let $f_n(x)$ be linear for $x\in [\frac {1}{n+1},K_n].$
We have $\|f_n\|=1+\frac {1}{2(n+1)}.$ 
For $x\in [\frac {1}{n+1},1]$ we have $(Tf_n)(x)\geq (Tf_n)(\frac {1}{n+1})=1.$ $$\text {So }\quad \|Tf_n\|\geq \int_{1/(n+1)}^1 (Tf_n)(x)dx\geq \int_{1/(n+1)}^1 1\cdot dx=$$ $$=1-\frac {1}{n+1}.$$
$$\text {So} \quad \frac {\|Tf_n\|}{\|f_n\|}\geq \frac {1-\frac {1}{n+1}}{   1+\frac {1}{2(n+1) }}.$$ 
A: 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 $$\frac{||\mathcal{T}f_n||}{||f_n||} \to 1\quad \text{as } n \to \infty.$$
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!
