If $f \in \mathcal{L}^{2}(\mathbb{R}^{n})$, does it imply that it is bounded almost everywhere?

Question

If $f \in \mathcal{L}^{2}(\mathbb{R}^{n})$, does it imply that it is bounded almost everywhere?

Answer 1

Consider the function $f : \mathbb{R} \to \mathbb{R}$,

$$f(x) = \begin{cases} x^{-1/3} & 0 < |x| < 1\\ 0 & \text{otherwise}. \end{cases}$$

Note that

$$\int_{\mathbb{R}}|f|^2dm = 2\int_0^1x^{-2/3} = 2\left[3x^{1/3}\right]_0^1 = 6 < \infty,$$

so $f \in L^2(\mathbb{R})$.

Now let $M > 0$ and let $K = \sqrt[3]{M}$. Note that $\{x \in \mathbb{R} \mid |f(x)| > M\} = \left(-\frac{1}{K}, \frac{1}{K}\right)\setminus\{0\}$ which has positive measure. Therefore, $f$ is not bounded almost everywhere. In particular, the answer to your question is no.

Answer 2

Here is one of my favorites, $$f(x) = \begin{cases} n & n < x < n+\frac{1}{n^4},\, n\in\mathbb{Z}^+\\ 0 & \text{otherwise}. \end{cases}$$ Then $$\limsup_{x\to\infty}f(x)=+\infty$$

while $$\int_\mathbb{R}|f(x)|^2\,dx=\frac{\pi^2}{3}$$