Let $E = C[a,b]$ and consider the max norm. I must prove that the operator $T:(E,\|f\|_0) \rightarrow (E,\|f\|_0)$ defined by $$T(f)(x) = \int_a^x f(t) \,dt$$ is uniformly continuous. My attempt was to show that this is a contraction. I've tryed to do something like
\begin{equation} \begin{split} |T(f)(x)-T(f)(y)| &= \left|\int_a^x f(t)\,dt - \int_a^y f(t) \,dt \right| \\ & \leq \int_y^x |f(t)| \, dt \\ & \leq \max_{x\in[a,b]} f(x) |x-y| \end{split} \end{equation}
But I don't know if this is right. Any help is welcome!
You are supposed to show that for any $\epsilon >0$ there exists $\delta>0$ such that $\|Tf-Tg||_{0} <\epsilon$ whenever $\|f-g\|_{0} <\delta$. So look at $|\int_a^{x}f(t)dt-\int_a^{x}g(t)dt|$. This is bounded by $\|f-g\|_{0}(b-a)$ for any $x$. Taking sup over $x$ we get $\|Tf-Tg||_{0}\leq \|f-g\|_{0}(b-a)<\epsilon$ if $\|f-g\|_{0} <\frac {\epsilon} {b-a}$. Take $\delta =\frac {\epsilon} {b-a}$.