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It is stated in my professor's notes that, given a sequence $\{f_j\}$ of $C_0^\infty(\Omega)$ functions (infinitely differentiable with compact support), and a function $g\in C_0^\infty(\Omega)$, all defined in an open set $\Omega\in\mathbb{R}^n$:

$$\|f_j-g\|_{L_p(\Omega)}\to0\implies|f_j(x)-g(x)|\to0\,\forall x\in\Omega$$

I was not able to prove it, however. Searching on the Internet, I found many counterexamples to the statement that $L^p$-convergence implies pointwise convergence, but they do not deal with $C_0^\infty(\Omega)$ functions, so it may also be true. Can anyone point to a proof or a counterexample?

Svetoslav
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Fiat Lux
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    Take an $f \in C_c^\infty$ with $f(0) \neq 0$ and set $f_j(x) = f(j\cdot x)$. Then $f_j \to 0$ in $L^p(\mathbb{R}^n)$, but not pointwise. However, this sequence converges pointwise almost everywhere, and for any convergent sequence in $L^p$, you have a subsequence that converges pointwise almost everywhere. It could be that with smooth functions, the full sequence converges pointwise almost everywhere, I don't see a counterexample to that right now. – Daniel Fischer Sep 25 '13 at 00:33
  • No, you don't have pointwise a.e. convergence of the full sequence for smooth sequences with smooth limits either. – Daniel Fischer Sep 25 '13 at 00:45
  • @DanielFischer, thank you, if you convert your first comment into an answer I will accept it – Fiat Lux Oct 14 '13 at 11:44

1 Answers1

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The assertion is incorrect. $L^p$-convergence (for $p < \infty$) does not imply pointwise convergence even for compactly supported smooth functions.

An easy example is a shrinking bump function, if $f \in C_c^\infty(\mathbb{R}^n)$ has $f(0) \neq 0$, then the sequence $(f_j)$ where $f_j(x) = f(j\cdot x)$ converges to $0$ in $L^p$, but $f_j(0) = f(0)$ for all $j$.

You don't even have pointwise convergence almost everywhere, as was the case in the above example. If you take a compactly supported function $f$ that takes the value $1$ on the entire unit hypercube (whichever dimension the space has), and then shrink the support by scaling with $\frac1k$, then moving around the scaled function so the value-$1$ plateau covers the entire hypercube (thus you first have $2^d$ translates of the scaled-by-$\frac12$ function, followed by $3^d$ translates of the scaled-by-$\frac13$ function, then $4^d$ translates ...) before the next shrinking, you have a sequence that does converge to $0$ in $L^p$, but not in any point of the unit hypercube.

However, $L^p$ convergence always (smooth or not) implies the existence of a subsequence that converges pointwise almost everywhere to the $L^p$ limit function.

Daniel Fischer
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