Consider $\phi: C([a,b]) \to R$ given by $$f \to \int_{a}^ b f dx$$ Show that $\phi$ is uniformly continuous.
Not really sure where to start here. It seems like I'm given very little. I'm given information about f, not $\phi$, and $f$ is not equal to that integral, it is mapped to there because of the arrow. So I'm not really sure what to do with what is given. Any help is appreciated!
The inequality
$$\vert \phi(f) - \phi (g) \vert =\left\vert \int_a^b \ f(x)\ dx - \int_a^b \ g(x)\ dx \right\vert \le (b-a) \Vert f - g \Vert$$
where $$ \Vert f-g\Vert= \sup_{x \in [a,b]} \vert f(x) -g(x) \vert$$
proves that $\phi$ is $(b-a)$-Lipschitz and is sufficient to get your desired conclusion. Providing that $\mathcal C([a,b], \mathbb R)$ is equipped with the $\sup$ norm.