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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?

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marked as duplicate by Nate Eldredge, njguliyev, Arthur, Nick Peterson, azimut Oct 27 '13 at 15:43

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    $\begingroup$ 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. $\endgroup$ – Zarrax Oct 27 '13 at 16:01
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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 $L^2(\mathbb R)$.

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