How to prove that following map is discontinuous?

Question

Consider the projection operator $$P : W^{1,2}(U) \to L^{\infty}(U)$$ given by $$P(f) = \chi \{f \geq 0\}$$ where

1. $\chi$ is characteristic function.
2. $W^{1,2}(U)$ is sobolev space on a bounded domain $U \subset \mathbb{R}^n.$
3. $\{f \geq 0\} = \{ x \in U: f(x) \geq 0\}$.

I have to prove that the defined map $P$ is discontinuous.

How to prove this?

Answer 1

This map is as discontinuous as it gets.

If you're looking for the simplest example possible, look at the constant functions $f = 0$ and $f_k = -1/k$. Then $P(f) = 1$ and $P(f_k) = 0$ for all $k$.

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Note that for two measureable subsets $A,B \subseteq U$, the $L^\infty$-distance between $\chi_A$ and $\chi_B$ is either $0$ (if $A=B$ up to a set of measure zero) or $1$ otherwise.

If $P$ is continuous at point $f \in W^{1,2}(U)$, then $P$ has to be constant on some neighborhood of $f$. If $P$ was continuous everywhere, it would be constant on the whole of $W^{1,2}(U)$, since this space is connected. And of course it isn't constant.

Actually, this map is everywhere discontinuous for $n > 1$. This fact follows from the existence of unbounded functions in $W^{1,2}(U)$. Let me know in the comments if you need an explanation.

Answer 2

Hint: Find a sequence of functions $\{f_n\} \subset W^{1,2}(U)$ with $f_n \ge 0$ on some set of positive measure, but $f_n \to f$ with $f < 0$ a.e.