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?

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    $\begingroup$ Why is $G_n$ closed? $\endgroup$ Commented Oct 27, 2020 at 9:30
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    $\begingroup$ You also seem to be confusing yourself. The idea is to prove that the $G_n$ have empty interior. $\endgroup$ Commented Oct 27, 2020 at 9:31
  • $\begingroup$ @WoolierThanThou Oh, that's right. I edit that. $\endgroup$
    – ferhenk
    Commented Oct 27, 2020 at 9:32
  • $\begingroup$ 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 $\endgroup$ Commented Oct 27, 2020 at 12:00
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    $\begingroup$ @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. $\endgroup$ Commented Oct 28, 2020 at 6:41

1 Answer 1


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.

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

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