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How to prove that the intersection of $L^1(\mathbb{R})$ space and $L^2(\mathbb{R})$ space is dense in $L^2(\mathbb{R})$ space?

Nate Eldredge
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user103577
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    The answer to that one proves a much more general statement, so I thought it might be instructive to see a concrete argument for the question at hand. – Zarrax Oct 27 '13 at 16:01

1 Answers1

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If $f \in L^2(\mathbb R)$, then define $f_n(x)$ by $f_n(x) = f(x)$ if $|x| < n$ and $f_n(x) = 0$ if $|x| \geq n$.

Then $f_n(x) \in L^1(\mathbb R) \cap L^2(\mathbb R)$ and $\lim_{n \rightarrow \infty} (f(x) - f_n(x)) = 0$ pointwise. So by the dominated convergence theorem $$\lim_{n \rightarrow \infty} \int_{\mathbb R}|f-f_n|^2 = 0 \tag1$$ The dominating function here is just $|f|^2$, since $|f-f_n| \leq |f|$.

Equation $(1)$ is equivalent to saying $\lim_{n \rightarrow \infty} ||f-f_n||_{L^2}^2 = 0$, so we have that $\lim_{n \rightarrow \infty} ||f-f_n||_{L^2} = 0$ too.

So we conclude that $L^1(\mathbb R) \cap L^2(\mathbb R)$ is dense in $L^2(\mathbb R)$.

Zarrax
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  • I know this is old but I cant see how $f$ is in the intersection. It is clear that it is in L^2 but I'm unable to see why it is in L^1. – tangentbundle Aug 17 '20 at 06:02
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    $f$ is not in the intersection. It's an arbitrary member of $L^2$ and I'm showing that $f_n \in L^1 \cap L^2$ converge to $f$ in the $L^2$ norm as $n$ goes to infinity. That is what density means here. – Zarrax Aug 20 '20 at 18:40
  • Oh I'm sorry, I meant $f_n$. Sorry about that. – tangentbundle Aug 24 '20 at 21:08
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    It's because by Holder's inequality or the Cauchy-Schwartz inequality, $||f_n||{L^1} = \int{|x| < n}|f(x)| * 1 \leq (\int_{|x| < n} |f(x)|^2)^{1 \over 2} (\int_{|x| < n} 1^2)^{1 \over 2}$ is finite since $f \in L^2(R^n)$. – Zarrax Sep 06 '20 at 02:03