I'm currently working on the following:
Define the function $\psi: C[a,b] \to \mathbb{R}$ by $\begin{equation*} \psi(f)=\int_{a}^{b}f(x)\,dx \end{equation*}$ for each $f \in C[a,b]$.
Show that $\psi$ is Lipschitz on $C[a,b]$ with the metric induced by the maximum norm $\Vert \cdot \Vert_{\infty}$.
So, what we want to show is $\Vert \psi(f) -\psi(g)\Vert_{\infty} \leq c\Vert f-g \Vert_{\infty}$.
Now, $\Vert \psi(f) – \psi(g) \Vert_{\infty}=\Vert \int_{a}^{b}f(x)dx – \int_{a}^{b}g(x)dx\Vert_{\infty} = \Vert \int_{a}^{b}f(x)-g(x)\,dx\Vert_{\infty}$
And this is about as far as I got. Maybe it's working with the maximum norms, but I am having great difficulty figuring out how to turn this into something resembling $\leq c \Vert f-g \Vert_{\infty}$.
I'm wondering if I can use uniform continuity to help me, since $f$ is mapping a compact metric space into a metric space.
Please Help!
Best Answer
You are mixing stuff up.
The functional $\psi$ takes values in $\mathbb{R}$, so start with $|\psi(f)-\psi(g)|$ and use the fact that $|\int h| \le \int |h|$.
Then use the fact that $|h(x)| \le \|h\|_\infty$ for all $x$.
In the above, $h(x) = f(x)-g(x)$.
Explicitly:
\begin{eqnarray} |\psi(f)-\psi(g)| &=& |\int_a^b (f(x)-g(x)) dx | \\ &\le& \int_a^b |f(x)-g(x)| dx \\ &\le& \int_a^b \| f-g\|_\infty dx \\ &=& (b-a) \| f-g\|_\infty \end{eqnarray}