# what functions or classes of functions are $L^1$ but not $L^2$

We know that if a real function is in $L^2$ then it is in $L^1$, but the reverse is not necessarily true. So what are the examples of functions that are $L^1$ but not $L^2$, especially those intrinsically unbounded ones if any? Welcome more examples, or classes of examples. Thanks.

I assume that you mean integrable functions in $[0,1]$.

In the sense of Baire almost every function in $L^{1}[0,1]$ is not in $L^{2}[0,1]$:

The space $L^{2}[0,1]$ is meager in $L^{1}[0,1]$ (that is to say it is a countable union of sets whose closure has empty interior in $L^{1}$).

The easiest way to see this by using the open mapping theorem: the inclusion $L^{2}[0,1] \to L^{1}[0,1]$ is continuous but not onto. More explicitly, the set $B_{n} = \{f \in L^{1}\,:\,\int |f|^{2} \leq n\}$ is easily seen to be closed and have empty interior and $L^{2}[0,1] = \bigcup_{n=1}^{\infty} B_{n}$. Similarly, if $1 \leq p < q \leq \infty$ then $L^{q}[0,1] \subset L^{p}[0,1]$ is meager.

This result is analogous to Baire's theorem saying that almost every continuous function on $[0,1]$ is nowhere differentiable, and with the same defect: If you choose a 'generic' function it won't be differentiable (or square-integrable) but from the statement you don't have a clue what such a function looks like.

• I know this was a long time ago but could you explain a bit more of how the Open Mapping Theorem shows us that the sets $B_n$ have no interior? – inkievoyd Jul 28 '18 at 2:59

Neither containment holds in general. On $\mathbb{R}$, the function $f$ such that $f(x)=1/x$ if $x\geq1$ and $f(x)=0$ otherwise is in $L^2\setminus L^1$, and the function $g$ such that $g(x)=1/\sqrt{x}$ if $0\lt x\leq1$ and $g(x)=0$ otherwise is in $L^1\setminus L^2$.

For bounded domains, $L^2\subset L^1$, because what keeps a function from being integrable on a bounded set is being too large, and squares make large numbers larger. If $f$ is a bounded function in $L^1$, then $f$ is also in $L^2$, because what keeps a bounded function from being integrable is not going to zero fast enough, and squares make numbers go to zero faster.

• +1. I think your reply is really nice! Partially answer my question – Tim Dec 28 '12 at 18:53
• Why is g(x) measerable? – Tobias Molenaar Dec 27 '17 at 12:23
• @TobiasMolenaar It is continuous on the given domain. – EA304GT Apr 6 '18 at 17:47

For an important example,

Check out Kolmogorov's example of a function in $L^1$ whose Fourier series diverges almost everywhere: http://books.google.com/books?id=ikN59GkYJKIC&pg=PA1

By Carleson's theorem this cannot be in $L^2$.

• But fortunately you don't need such heavy artillery to see that (non-square integrability of that example, I mean). – t.b. Jan 21 '11 at 4:46
• Something for mathoverflow.net/questions/42512/… perhaps? ;) – Hans Lundmark Jan 21 '11 at 11:36
• @Hans: Perhaps :-) Liked the statement "FLT is not strong enough to prove $\sqrt{2}$ is irrational" :-) But Carleson's theorem is an important result and I thought it was worth mentioning. Besides OP did ask for a class of examples :-) – Aryabhata Jan 21 '11 at 17:47
• It seems like Kolmogorov's result can be improved to everywhere divergence. Anyone has the example of everywhere divergence. – Rachid Atmai Oct 30 '19 at 22:47

You can get an example from any two sequences of positive reals such that $\sum \alpha_n \beta_n < \infty$ but $\sum \alpha_n^2 \beta_n = \infty$, e.g. $\alpha_n = n$ and $\beta_n = 1/n^3$.

For the reverse non-containment, take $\alpha_n = 1/n$ and $\beta_n = 1$.

If we look at the discrete group $$\mathbb{Z}$$, we have $$L^1(\mathbb{Z})=\ell^1(\mathbb{Z})\subset \ell^2(\mathbb{Z})=L^2(\mathbb{Z})$$ - see the thread How do you show that $$\ell^p\subset \ell^q$$ for $$p≤q$$?.

If we look at the compact group $$\mathbb{T}$$ (the unit circle), we have $$L^2(\mathbb{T})\subset L^1(\mathbb{T})$$, since by Hölder's inequality $$\|f\|_{L^1}=\int|f|dx=\int1\cdot|f|dx\leq\sqrt{\int1^2dx}\sqrt{\int|f|^2dx}=1\cdot\|f\|_{L^2}=\|f\|_{L^2}$$

If we look at the continuous non-compact group $$\mathbb{R}$$ there is no inclusion.

• that containment is reversed in second paragragh – Believer Oct 1 '19 at 15:22

Coming across these answers, it might be valuable to add the following explicit calculation:

Let $$f \in L^1(\mathbb{R}^d) \cap L^{\infty}(\mathbb{R}^d)$$. Then $$f \in L^2(\mathbb{R}^d)$$.

Indeed, denoting by $$dx$$ the Lebesgue measure on $$\mathbb{R}^d$$, we have $$\int_{\mathbb{R}^d}|f|^2 = \int_{|f| \geq 1} |f|^2 + \int_{|f| < 1} |f|^2 \leq \int_{|f| \geq 1} C^2 + \int_{|f| < 1}|f| \leq C^2 \cdot dx(|f| \geq 1) + ||f||_{L^1} < +\infty.$$ The first inequality uses the essential boundedness of $$f$$ and the last inequality uses $$f \in L^1(\mathbb{R}^d),$$ which yields $$dx(|f| \geq 1) < +\infty$$.

On an unbounded domain $$D$$ (such as $$D= \mathbb{R}^d$$), the "converse", i.e. $$L^{\infty}(D) \cap L^2(D) \subseteq L^1(D)$$ does not hold: While we can estimate $$\int_D|f| \leq \int_{D \cap |f| > 1} |f|^2+\int_{D \cap |f| \leq 1} |f| \leq ||f||_{L^2(D)}+dx(D\cap \{|f| \leq 1\}),$$we cannot conclude $$f \in L^1(D)$$, because the last summand is certainly not finite. As nicely explained in the above answers, this may be understood as insufficient fast decay to $$0$$ of $$|f|$$ on $$D \cap \{|f| \leq 1\}$$, while the decay of $$|f|^2$$ on the same set is fast enough.