If $f$ is a continuous function $[0,1]\to \mathbb R$, we define a linear application $T$ as follows $$T(f)(x)=\begin{cases} f(0) & \mathrm{if }~ x=0 \\[0.2cm] \displaystyle \frac{1}{x}\int_0^xf(t)dt & \mathrm{if }~ 0<x\leqslant 1\end{cases}$$
Intuitively it seems that for $0 \leqslant x \leqslant 1$, $$\displaystyle\lim_{{n\to\infty}}T^n(f)(x)= f(0)$$
Do we have $$|| T^n(f)-f(0)||_{\infty}\longrightarrow 0$$ as $n\to\infty$?
Yes. By induction, $T^n(f)(x) = \frac{1}{n!x} \int_0^x \log^n\left(\frac{x}{u}\right) f(u)\, du$ for $0 \le x < 1$. The formula can also be written $$T^n(f)(x) = \frac{1}{n!}\int_0^1 \log^n\left(\frac{1}{u}\right) f(ux)\, du.$$
If $f = 1$, then $T^n(f) = 1$ for all $n$ and thus $\|T^n(f) - f(0)\|_\infty = 0$ for all $n$. When $f(x) = x^m$ for some $m \ge 1$, $T^n(f)(0) = 0$ and for $0 < x \le 1$, $T^n(f)(x) = \frac{x^m}{(m+1)^{n+1}}$. Thus $\|T^n(f) - f(0)\|_\infty \to 0$ as $n\to \infty$. By linearity, your conjecture is true when $f$ is a polynomial. The general case now follows from Weierstrass's approximation theorem and the estimates $\|T^n(f)\| \le \|f\|$, for all $n$. Indeed, given $\varepsilon > 0$, choose a polynomial $p$ such that $\|f - p\|_\infty < \frac{\varepsilon}{3}$. Choose a positive integer $N$ such that $\|T^n(p) - p(0)\|_\infty < \frac{\varepsilon}{3}$ for all $n \ge N$. So if $n \ge N$,
\begin{align} \|T^n(f) - f(0)\|_\infty &\le \|T^n(f) - T^n(p)\|_\infty + \|T^n(p) - p(0)\|_\infty + |p(0) - f(0)|\\ &= \|T^n(f - g)\|_\infty + \|T^n(p) - p(0)\|_\infty + |p(0) - f(0)|\\ &\le 2\|f - g\|_{\infty} + \|T^n(f) - f(0)\|_{\infty}\\ &< \frac{2\varepsilon}{3} + \frac{\varepsilon}{3}\\ &= \varepsilon \end{align}
Since $\varepsilon$ is arbitrary, $\|T^n(f) - f(0)\|_\infty \to 0$ as $n\to \infty$.