Is there a smooth function, which is in $L^1$, but not in$L^2$? [closed]

Question

I am studying measure theory. While going over $L^p$-spaces I asked myself, whether there is $f\in C^\infty(\mathbb{R})$ s.t. $f\in L^1(\mathbb{R})\setminus L^2(\mathbb{R})$? I assume there could be some kind of density argument, but I am not sure.

Answer 1

Rationale: the typical example of a function in $L^1(\mathbb{R}) \setminus L^2(\mathbb{R})$ is $\frac{1}{\sqrt{x}}$ on $(0, 1)$ and $0$ elsewhere. Unfortunately this function is clearly not smooth or even continuous. But what exactly makes it belong to $L^1$ but not $L^2$? It is not any sort of local quality like smoothness, but rather the magnitude of its values and the measure of the sets where these values are attained. So maybe we can "rearrange" the function so that it retains its magnitude while becoming smooth. Since a smooth function must be bounded on any bounded interval, we should diffuse the unboundedness over an unbounded interval.

The concrete example goes as follows: imagine a function $g$ such that for every $n \in \mathbb{N}$, the graph of $g$ around the point $n$ looks like a rectangle of width $\frac{1}{n^2}$ and height $\sqrt{n}$, and the function is zero elsewhere. Formally,

$$g(x) = \sum_{n=1}^{\infty} \sqrt{n} \cdot \chi_{\left[ n-\frac{1}{2n^2}, n+\frac{1}{2n^2} \right]}(x).$$

It is easy to see that $g \in L^1(\mathbb{R}) \setminus L^2(\mathbb{R})$, because the series $\sum_{n=1}^{\infty} \frac{\sqrt{n}}{n^2}$ converges while the series $\sum_{n=1}^{\infty} \frac{n}{n^2}$ diverges. Now it suffices to make $g$ smooth without significantly changing its shape, for example by replacing the discontinuities with adequately steep smooth functions, as described here.

Answer 2

We note that $C^\infty_c((0, 1)) \subset L^2((0, 1))$ is dense. (Note that I’m using the open interval $(0, 1)$, so compactly supported means supported on some $[a, b]$ with $0 < a < b < 1$.) Now, for any $n \geq 0$, the function $g_n = 1_{(0, 2^{-2(n + 1)})}$ satisfies the condition that

$$\|g_n\|_1 = 2^{-2(n + 1)} < 2^{-n} \cdot 2^{-(n + 1)} = 2^{-n}\|g_n\|_2$$

Thus, by using density of $C^\infty_c((0, 1)) \subset L^2((0, 1))$ (and noting that $1$-norm is bounded by $2$-norm on $(0, 1)$), we see that there exists $f_n \in C_c^\infty((0, 1))$ s.t. $\|f_n\|_1 < 2^{-n}\|f_n\|_2$. By normalizing, we may assume $\|f_n\|_2 = 1$. Now define,

$$f(x) = \begin{cases} f_n(x - n)&, \; \text{if} \; n < x < n + 1 \; \text{for some} \; n \geq 0\\ 0&, \; \text{otherwise} \end{cases}$$

Then $f \in C^\infty(\mathbb{R})$ and,

$$\begin{split} &\|f\|_1 = \sum_{n \geq 0} \|f_n\|_1 \leq \sum_{n \geq 0} 2^{-n} < \infty\\ &\|f\|_2^2 = \sum_{n \geq 0} \|f_n\|_2^2 = \sum_{n \geq 0} 1 = \infty \end{split}$$

That is, $f \in L^1(\mathbb{R}) \setminus L^2(\mathbb{R})$.