$C([0,1])$ is the set of continuous functions $u:[0,1]\to \mathbb{R}.$ For any $u\in C([0,1])$ let $\displaystyle Lu(t) = \frac{1}{2}\int_0^1 e^{-u(x)}(x+t)\,dx.$ I'm to show that $L$ defines a mapping $L:C([0,1])\to C([0,1])$.
I want to show that for any $u\in C([0,1]),\; (Lu)$ is continuous.
\begin{align*} |Lu(t)-Lu(t')| &= \left|\frac{1}{2}\int_0^1e^{-u(x)}(x+t)\,dx -\frac{1}{2}\int_0^1e^{-u(x)}(x+t')\,dx \,\right| \\ &=\left|\frac{1}{2}\int_0^1e^{-u(x)}(t-t')\,dx\,\right|\\ &\leq \frac12 \, |t-t'| \cdot 2M \leq |t-t'|, \end{align*} where $M = 1$ is an upper bound on $e^{-u(x)}$.
Have I now shown that $L$ is (Lipschitz) continuous?
Why is $M=1$ an upper bound for $e^{-u}$ ??
With your proof we get:
$|Lu(t)-Lu(t')| \le \frac{1}{2}||e^{-u}||_{\infty}|t-t'|$.