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I want to ask about example of real valued functions defined on the real line such that their convolution exist in every point and is discontinuous on a "large" set, for example on each point of some interval or in dense subset of $\mathbb{R}$ or maybe on the whole $\mathbb{R}$.

My question is related to paper Mikusinski, Ryll-Nardzewski, Sur le produit de composition, http://matwbn.icm.edu.pl/ksiazki/sm/sm12/sm1213.pdf

The authors consider convolution of integrable functions which are zero for $x\leq 0$. On page 52 the above paper is given example of two integrable functions such that their convolution is discontinuous at a point. Autors say also, but there is no proof of this statement, that it is possible by condensation of singularities to construct integrable functions such that their product is discontinuous everywhere (on $\mathbb{R_+}$).

Thanks.

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I would try to consider $f*f$, where $f$ is the characteristic function of a fat Cantor set. –  Jack D'Aurizio Oct 13 '13 at 16:06
    
Related: mathoverflow.net/questions/92748/… –  Grigory M Oct 14 '13 at 10:11
2  
@JackD'Aurizio: The shift group action on $\mathbb{R}$ is strongly continuous in $L^1$, so the convolution of an $L^1$ function and a bounded function is always continuous. –  Alexander Shamov Oct 14 '13 at 10:35
    
Second try: consider the standard Cantor set $K\subset[0,1]$. Every $x\in[0,1]\setminus K$ belong to a maximal interval $(x^-,x^+)\subset[0,1]\setminus K$: define $f$ on $[0,1]$ such that $f(x)=0$ for any $x\in K$ and $f(x)=\exp\left(-\log(3/2)\log(x^{+}-x^-)/\log(3)\right)$ for any $x\in[0,1]\setminus K$. Then $f\in\mathcal{L}^1([0,1])$, unbounded in a right neighbourhood of zero, looks like a good candidate in order to ensure that $f*f$ is discontinous over a "large" set. –  Jack D'Aurizio Oct 14 '13 at 13:03
    
@JackD'Aurizio: Well, it's easy to construct an example of $L^1$ functions with colvolution that is everywhere locally unbounded, but the the tricky thing here is the condition that the convolution be "everywhere defined". I interpret it as absolute convergence of the relevant integral, which seems to fail in your example (see what happens when the right neighborhood of zero meets the left neighborhood of one). On the other hand, in the paper cited by Richard the singularity is in the right neighborhood for both functions, so this doesn't happen. –  Alexander Shamov Oct 14 '13 at 13:40
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1 Answer

up vote 12 down vote accepted
+500

Take $$f\left(x\right)=\begin{cases} x^{-1/2}, & x\in\left(0,1\right]\\ 0, & \text{otherwise} \end{cases} $$ This function is obviously in $L^1$; note also that $f\ast f$ is $0$ on $(-\infty,0] \cup [2,+\infty)$, $\pi$ on $(0,1]$, and it decays from $\pi$ to $0$ continuously on $[1,2]$. Therefore, $f \ast f(x)$ is everywhere defined, in the sense that $\intop |f(y) f(x-y)| dy < \infty$ for all $x$; it is bounded in $x$, and its only discontinuity is at zero.

Now take $f_a(x) := f(x-a)$, let $\{a_n\}$ be an arbitrary dense countable set, and obverve that:

  1. $F := \sum_n 2^{-n} f_{a_n} \in L^1$
  2. $f \ast F = \sum_n f \ast 2^{-n} f_{a_n}$ pointwise, by monotone convergence.
  3. Moreover, the series converges absolutely in $L^\infty$ norm, thus the limit is continuous outside $\{a_n\}$, and discontinuous at $\{a_n\}$.

Thus we have constructed a convolution which is everywhere defined and discontinuous at a dense countable set.

Now a general remark. Convolution, whenever it is defined, is a limit of continuous functions (by approximation of one of your $L^1$ functions by bounded ones), so it must be of Baire class one. In particular, if it's locally bounded then it must be almost everywhere continuous in the sense of category, so you cannot do much better than in my example.

Now assume that you have a convolution of $f,g \ge 0$. $f \ast g$ is still of Baire class one, but now with values in the extended real line $[0,\infty]$. So if $f \ast g$ is locally unbounded at every point of some interval, then since it must be Baire almost everywhere continuous (again, in the sense of the topology of $[0,\infty]$), there will be a point on that interval where $f \ast g$ is actually infinite, which probably counts as "nonexistence of convolution at that point".

The general case reduces to the case of positive functions: if your convolution exists everywhere, then it would also exist everywhere if we replaced $f,g$ by $|f|,|g|$.

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Really insightful, 500 well-deserved points. –  Jack D'Aurizio Oct 14 '13 at 15:33
    
+1 Shouldn't the first line be $x\in (0,1]$ ? –  leonbloy Oct 14 '13 at 15:40
    
@leonbloy: Yeah, especially since most of my answer is all about treating infinite values as bad... :) –  Alexander Shamov Oct 14 '13 at 15:45
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