Is the set of continuous function that are also $\mathcal{L}^1(\mathbb{R})$ a subset of $\mathcal{L}^2(\mathbb{R})$

Question

I have this problem, and I don't know what to do... I don't find a counterexample and I start thinking that this fact could be true... But the only idea I have to prove this fact is that $f$ must converge to $0$ as $x\to\infty,$ and I don't see a way to go on...

Answer 1

Inspired by @N. S., let $\phi\in C_{0}^{\infty}({\bf{R}})$ be such that $0\leq\phi\leq 1$, $\text{supp}(\phi)\subseteq\{|x|<1\}$, $\phi=1$ on $\{|x|<1/2\}$, and $f(x)=\displaystyle\sum_{n=1}^{\infty}n^{3/2}\chi_{I_{n}}(x)\phi(4n^{2}(x-n))$, where $I_{n}=[n-1/(4n^{3}),n+1/(4n^{3})]$, then $f\in C^{\infty}({\bf{R}})\cap L^{1}({\bf{R}})$ but $f\notin L^{2}({\bf{R}})$.

Answer 2

Pick $f(x)$ to be the step function $=n\sqrt{n}$ for $x \in [n-\frac{1}{n^2}, n+\frac{1}{n^2}]$ and then a fast drop to $0$, to make it continuous.

Then $f(x)$ is continuous, $L^1$, but $f$ is not $L^2$.

Note The claim would be correct if $f$ would be bounded or uniformly continuous.