The behavior of an $L^2(\mathbb{R})$ function close to $x=0$.

Question

Is there a way to prove that an $L^2(\mathbb{R})$ function close to $x=0$ is $\mathcal{O}(x^{-1/2})$? In other words, can we prove that there is a function $\tilde{f}$ equal to $f$ almost everywhere such that $$ \limsup_{x\to 0} \left(|x|^{1/2} |\tilde{f}|\right)<\infty? $$ I am working on a proof involving $L^2(\mathbb{R})$, and this fact would be really useful. It seems reasonable since the function needs to be integrable across $x=0$ but I am not exactly sure how to prove that. No need of the detailed proof, just an idea of what tool I should be using.

Answer 1

No.

One could give a formula for a counterexample in terms of powers and logs and such.That only works on the line - I prefer a more measure-theoretic construction that can be imitated elsewhere:

Choose $\delta_n\in(0,n^{-4}]$ small enough that if we say $$I_n=\left[\frac1{n^3},\frac1{n^3}+\delta_n\right]$$then the $I_n$ are disjoint. Let $$f=\sum_{n=1}^\infty n\chi_{I_n}.$$Then $f(t)$ is not $O(t^{-1/2})$. But $$||f||_2^2=\sum n^2\delta_n<\infty.$$

A comment mentions a generalization that you can prove using the same sort of construction:

Exercise. Suppose $g:(0,\infty)\to(0,\infty)$ and $\lim_{x\to0}g(x)=\infty$. There exist $f\in L^2((0,\infty))$ and $x_n\to0$ with $f(x_n)\ge ng(x_n)$ for all $n$.

Answer 2

No, here is a counterexample: let $f(x) = n$ on $[\frac{1}{n}, \frac{1}{n} + \frac{1}{n^3}]$, for all $n \in\mathbb{N}$ and 0 elsewhere. Then $f\not\in O(x^{-\frac{1}{2}})$ but $\int |f|^2 = \sum_n n \frac{1}{n^3} < \infty$, thus $f \in L^2$.