$V\colon \mathcal{B}(X, \mathbb{R}) \rightarrow \mathcal{B}(X, \mathbb{R})$, $V(f) = |f|$ is continuous

Question

Let $X$ be an arbitraty set and define $V\colon \mathcal{B}(X, \mathbb{R}) \rightarrow \mathcal{B}(X, \mathbb{R})$ as $V(f) = |f|$, where $|f|(x) = |f(x)|$. I need to show that $V$ is countinuous.

What I tried to do is showing that $V$ is a weak contraction map, ie, for every $f, g\colon X \rightarrow \mathbb{R}$, $d(V(f), V(g)) \leq d(f, g)$, but I am having some trouble with the argument.

One of the things I did was:

\begin{align*} d(V(f), V(g)) &= d(|f|, |g|)\\ &\leq d(|f|, f) + d(f, g) + d(g, |g|), \end{align*} by the triangle inequality.

But this upper bound doesn't seem to help me.

Could someone give me an insight?

Thanks a lot!

Answer 1

For any, $x$ and $y$ in a normed space, the triangle inequality says that $$|||x||-||y|||\le ||x-y||\le ||x||+||y||.$$

In particular, for any function $f$ and $g$ and any $x\in X$, you have $||f|(x)-|g|(x)|\le |f(x)-g(x)|$.

Taking the sup (which exists since everything is bounded) shows that $V$ is lipchitz and therefore continuous.