Prove that if $f \in C^1(E)$, where $E$ is a compact convex subset of $\mathbb R^n$, then $f$ satisfies a Lipschitz condition on $E$.
I have almost proved this entire statement but I'm kind of stuck at the very last step. I'm not entirely sure what to do to guarantee our constant is positive to satisfy the Lipschitz condition.
PROOF: Since $E$ is compact convex subset of $\mathbb R ^n$,
$$\begin{align} |f(x) - f(y)| &= \left|\int_0^1{d \over d\tau}f\left(y + \tau(x - y)\right)d\tau\right| \\ &\le \int_0^1 \left|{d \over d \tau} f\left(y + \tau(x - y)\right)\right|d\tau \\ & = \int_0^1|f'(y + \tau(x-y))(x-y)|d\tau\\ &\le \int_0^1\|f'(y + \tau(x-y))\||x - y|d\tau \\ &= \int_0^1\|f'(y + \tau(x-y))\|d\tau |x -y| \\ & \leq \;\;\;\;??? \\ & = K|x - y|\end{align}$$
Can anyone provide a hint on how to ensure the integrand in nonzero? I've looked in my textbook for a similar proof, but they only do locally Lipschitz. In that proof, they choose the max of the function on $E$, but how would I do that here?
So $f \in C^1(E)$. Hence, $f'$ exists and is continuous. By continuity of $f'$ and compactness of $E$, the function $f'$ is bounded on $E$, i.e. there exists a number $M$ such that $f' \leq M$ on $E$. Pick any two points $x$ and $y$. Because the line segment joining them lies inside $E$, by convexity, the mean value theorem applies i.e. there exists a point $z$ on the line joining $x$ and $y$ such that $f(x)-f(y) = f'(z)(x-y)$. Since $f'(z) < M$, we get that $f(x)-f(y) < M(x-y)$. Hence, we are done.