Value of operator norm when $\mathcal{T}f(x)=\int^{x}_{0} f(t)dt$

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 $(Tf_n)(x)\geq (Tf_n)(\frac {1}{n+1})=1.$ $$\text {So }\quad \|Tf_n\|\geq \int_{1/(n+1)}^1 (Tf_n)(x)dx\geq \int_{1/(n+1)}^1 1\cdot dx=$$ $$=1-\frac {1}{n+1}.$$

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