# Existence of $L^1((0,1))$ functions which blow up on every open interval

Consider an open interval $$(0,1) \subset \mathbb{R}$$ and the subset $$\mathcal{F} := \{f \in L^1((0,1)): \|{f\vert_{(a,b)}}\|_{\infty} = \infty \, \forall \, 0 \leq a < b < 1\} \subset L((0,1), dx).$$ I want to show that $$\mathcal{F}$$ is non-empty in $$(L((0,1), dx), \| \, \|_{L^1})$$.

For that reason, I define the sets $$G_n := \{f \in L^1((0,1)): \|{f\vert_{(a,b)}}\|_{\infty} \leq n \, \text{ for some } 0 \leq a < b < 1\}$$ and aim to show two things:

• $$G_n$$ is closed for every $$n \in \mathbb{N}$$
• $$G_n$$ has empty interior for $$n \in \mathbb{N}$$

Then, $$F_n := L^1{((0,1))}\setminus G_n$$ is open and dense for every $$n \in \mathbb{N}$$ and $$\mathcal{F} = \bigcap_{n \geq 1} F_n$$ is non-empty by Baire's category theorem (actually, it will be comeager).

Now, closedness of $$(G_n; n \geq 1)$$ is clear.

To prove that $$G_n$$ have empty interior for every $$n$$, I assume, for the sake of contradiction, that they have not. That is, let $$n \in \mathbb{N}$$. Then, for every $$g \in G_n$$, there exists $$\varepsilon > 0$$ s.t.

$$B := \{h \in L^{1}((0,1)): \|h - g \|_{L^1} < \varepsilon\} \subset G_n.$$ The idea is now to construct a function $$h \in L^1((0,1))$$ s.t. $$h \in B$$ and $$h \notin G_n$$, i.e. $$\|h-g\|_{L^1} < \varepsilon /2 \quad \text{ and} \quad \|h\vert_{(a,b)} \|_{\infty} > n \quad \forall \, 0 < a < b < 1.$$ Is this a feasible approach? If yes, can you hint how this may work?

• Why is $G_n$ closed? Commented Oct 27, 2020 at 9:30
• You also seem to be confusing yourself. The idea is to prove that the $G_n$ have empty interior. Commented Oct 27, 2020 at 9:31
• @WoolierThanThou Oh, that's right. I edit that. Commented Oct 27, 2020 at 9:32
• Do you guys believe that it is impossible to construct such a function by hand? I mean constructing it as a limit of a sequence, obviously Commented Oct 27, 2020 at 12:00
• @JustDroppedIn Sure, you can. Let $f_n=\sum_{k=0}^{2^{n-1}} 1_{[k2^{-n},k 2^{-n}+4^{-n})}$. Then, $\sum_{n} f_n$ is convergent in $L^1$ and it explodes at every dyadic rational point. Commented Oct 28, 2020 at 6:41

Okay, so it's not obvious (or even true) that $$G_n$$ is closed because the interval on which your bound holds might shrink. Indeed, let $$f_m=n 1_{(0,2^{-m})}+2n 1_{[2^{-m},1)}$$. Then, clearly, $$f_m\in G_n$$, but $$f_m\to 2n$$ in $$L^1$$.

Instead, define $$G_{n,m,k}=\{ f\in L^1((0,1))|\; \|f|_{(k2^{-m},(k+1)2^{-m})}\|_{\infty}\leq n\}$$ where $$k$$ ranges over $$\frac{1}{2}\{1,...,4^m-2\}$$. These $$G_{n,m,k}$$ are closed since any $$L^1$$-convergent sequence admits an almost everywhere convergent set of representatives (by passing to a subsequence, of course). This follows, since if $$f_{\alpha}$$ is a choice of representatives of a sequences in $$G_{n,m,k}$$ converging to $$f$$ pointwise we have, for every $$\varepsilon>0$$ $$\{|f|_{(k2^{-m},(k+1) 2^{-m})}|\geq n+\varepsilon\}\subseteq \bigcup_{N\in \mathbb{N}} \bigcap_{\alpha \geq N} \{|f_{\alpha}|_{k2^{-m},(k+1) 2^{-m}}| \geq n+\varepsilon/2\},$$ and the right-hand side is a null-set. Since $$\varepsilon$$ was arbitrary, we get that $$f$$ represents a class in $$G_{n,m,k}$$.

Now, I'll leave it to you to verify that, indeed, $$\cap_{n,m,k} L^1((0,1))\setminus G_{n,m,k}=\mathcal{F}$$.

Now, that $$G_{n,m,k}$$ is nowhere dense is easy. Indeed, let $$f\in G_{n,m,k}$$ be arbitrary and let $$\varepsilon>0$$. Then, let $$A=(k2^{-m},(k+\frac{\varepsilon}{1337n})2^{-m})$$ and define $$g:=f+1337n 1_A$$. Then, obviously $$\|g-f\|=1337n \cdot dx(A)=\varepsilon 2^{-m}$$. However, the essential supremum of $$g$$ over $$(k2^{-m},(k+1)2^{-m})$$ is easily seen to be at least $$1336n$$. We conclude that $$G_{n,m,k}$$ is nowhere dense, and you are done.

Edit: On the request of JustDroppedIn, let's also show that we can give a perfectly constructive proof of the existence of $$L^1$$-functions which blow up on every interval.

Let $$\phi_n=\sum_{k=0}^{2^n-1} 1_{[k2^{-n},k2^{-n}+4^{-n})}$$. Then, $$\|\phi_n\|_{L^1}=2^{-n}$$ and hence, $$\sum_{n=1}^{\infty}\phi_n$$ is absolutely convergent in $$L^1$$ and hence, defines an $$L^1$$ function. This function also clearly has the property that it blows up at every dyadic rational number and hence, in every interval. Of course, this construction admits a myriad of variations.

Second edit: I don't know why I thought that the constructive proof wouldn't give density. Indeed, the functions that blow up everywhere must either form a dense or empty subset of $$L^1$$. Just note that if $$f$$ is an $$L^1$$ function which blows up everywherehere, then $$\varepsilon f$$ is also an $$L^1$$ function with norm $$\varepsilon \|f\|_{L^1}$$ and it also has the property of blowing up everywhere. This implies that there are $$L^1$$ functions arbitrarily close to $$0$$ which blow up everywhere. But this implies density of the functions which blow up everywhere. What you don't get is the stronger conclusion of Baire, that these functions form a co-meagre set.

• huh, 1337=leet? Commented Oct 27, 2020 at 9:51
• Just having fun with my arbitrary choices of constants. :) Commented Oct 27, 2020 at 9:51
• cool cool cool cool Commented Oct 27, 2020 at 9:52