# Is there a useful relationship between pointwise and $L^2$ distance?

It would be really convenient to get a bound on the point-wise closeness of functions by knowing their $L^2$ distance. Clearly, if two functions are close in the $L^2$ sense, you cannot get a general bound on their point-wise distance, but maybe they are close along some parts of their domain.

In my mind a theorem might go something like this:

Given a measurable, bounded function $f:U\rightarrow\mathbb{R}$, if $||f||_2<\varepsilon$ then there is some 'small' $\delta > 0$ and a set $V\subset U$ where $|f(x)|<\delta \ \forall\ x\in V,$ and $\mu(V)\geq\mu(U)-\eta_{\varepsilon, \ \delta}$

Then if we had $f_1,f_2\ni||f_1-f_2||_2<\varepsilon$ there would be a 'fairly large' set $V$ where $|f_1-f_2| < \varepsilon$ throughout $V$.

Do theorems like this exist?

• Someone just noted that the $L^2$ norm is defined on equivalence classes of functions. However, I think we should still be able to say something about the measure of the domain where the functions are close, since the places '$L^2$-equivalent' functions differ has to be of measure 0. – Christian Chapman May 20 '14 at 5:15

The following is a naive take on this problem.

If $f: U \rightarrow \mathbb{R}$ is a measurable function such that $$\sqrt{\int_U f^2} < \epsilon_1$$

and $V \subset U$ is the measurable set such that $\left|f(x)\right| \geq \epsilon_2$ for all $x \in V$ and $\left|f(x)\right| < \epsilon_2$ for all $x \in U\setminus V$, then

$$\epsilon_2^2 m(V) = \int_{V}\epsilon_2^2 \leq \int_{U}f^2 < \epsilon_1^2$$

Hence $$m(V) < \frac{\epsilon_1^2}{\epsilon_2^2}$$

It's not difficult to construct simple functions that show that $m(V)$ can be arbitrarily close to this upper bound. Thus without additional conditions this is the best bound we can get on the size of $V$.

This suggests the result,

Proposition: Let $\epsilon > 0$. If $f:U \rightarrow \mathbb{R}$ is a measurable function such that $$\sqrt{\int_U f^2} < \epsilon$$ then the set $V = \left\{x \in U : \left|f(x)\right| \geq \sqrt{\epsilon}\right\}$ has measure $m(V) < \epsilon$.

Perhaps with additional conditions a stronger result can be proven.