# Prove that $\phi : C[a,b]\rightarrow \mathbb{R}$ given by $\phi(f)=\int_a^bfdx$ is uniformly continuous

Prove that $\phi : C[a,b]\rightarrow \mathbb{R}$ given by $\phi(f)=\int_a^bfdx$ is uniformly continuous.

First, since $f$ is continuous $\phi$ is well defined (integrals exist). Since $f$ is continuous on a compact set it is bounded, ie. $|f(x)|< M$ where $M>0$ is a constant.

Now fix $\epsilon>0$. If we take $\delta=\frac{\epsilon}{(b-a)}$ then $$|\phi(f)-\phi(g)|=\Bigg | \int_a^b (f(x)-g(x))dx \Bigg |\le \int_a^b |f(x)-g(x)|dx <\epsilon$$ whenever $f,g\in C[a,b]$ and $\sup_{x\in [a,b]} |f(x)-g(x)|<\delta$.

Is this clear enough proof or am I missing something? thanks.

• Yes. The procedure is "almost" correct.. To ensure the strictly inequality, take $\delta=\frac{\epsilon}{2(b-a)}$ – sinbadh Jan 3 '16 at 8:41
• Oh, good point. Thanks! – DiscoPotato Jan 3 '16 at 8:44