Showing continuity in supremum norm

Question

Let $C([0,1])$ be the set of continuous functions $u:[0,1]\to \mathbb{R}$.

For any $u\in C([0,1])$, let a map $(Lu)(t) = \frac{1}{2}\int\limits_0^1 e^{-u(x)}(x+t)dx$

Let the space $C([0,1])$ have the sup-norm metric, ie $d(u,v) = \underset{x\in[0,1]}{sup}\{|u(x)-v(x)|\}$

I'm to show that L is a continuous function with respect to the sup-norm metric.

So I tried letting $u,v \in C([0,1])$, then

$|(Lu)(t)-(Lv)(t)| = |\frac{1}{2}\int\limits_0^1 (x+t)[e^{-u(x)}-e^{-v(x)}]dx|$,

but I don't know how to proceed from here, the $(x+t)$-bit throws me.

Help?!

Answer 1

Hint: $|(Lu)(t)-(Lv)(t)| = |\frac{1}{2}\int\limits_0^1 (x+t)[e^{-u(x)}-e^{-v(x)}]dx| $

$\le \frac{1}{2} \int_0^1(|x|+|t|)|e^{-u(x)}-e^{-v(x)}| dx \le \int_0^1|e^{-u(x)}-e^{-v(x)}| dx $,

since

$|x|+|t| \le 2$ for $t,x \in [0,1]$.

Answer 2

We need to prove that for two functions $u$ and $v$ if $$\sup_{0<x<1}|u(x)-v(x)|<\epsilon$$which is equvalent to the following $$\forall 0<x<1 \qquad,\qquad |u(x)-v(x)|<\epsilon$$then$$\forall 0<t<1\qquad,\qquad|Lu(t)-Lv(t)|<k\epsilon$$with $k$ being a constant. Also by definition we have$$|Lu(t)-Lv(t)|=\frac{1}{2}|\int\limits_0^1 \left(e^{-u(x)}-e^{-v(x)}\right)(x+t)dx|\le\frac{1}{2}\int\limits_0^1 \left|e^{-u(x)}-e^{-v(x)}\right|(x+t)dx$$since both $u$ and $n$ have been defined on a compact set and are continuous so their range is a compact set and is bounded. Let$$|u(x)|<u$$therefore $$-u<u(x)<u\\-\epsilon<v(x)-u(x)<\epsilon$$substituting this in the latter integral gives us$$\frac{1}{2}\int\limits_0^1 \left|e^{-u(x)}-e^{-v(x)}\right|(x+t)dx\le\dfrac{e^u(e^{\epsilon}-1)}{2}\int\limits_0^1 x+tdx<e^u(e^\epsilon-1)$$using Taylor series$$e^u(e^\epsilon-1)=e^u(1+\epsilon+\dfrac{\epsilon^2}{2}+\dfrac{\epsilon^3}{6}+...-1)$$and for small enough $\epsilon$ we have $$e^u(e^\epsilon-1)<2e^u\epsilon=k\epsilon$$which is what we wanted to prove.