What theorem said that $L^2[a,b]$ is dense in $L^1[a,b]$ in Royden "Real analysis " 4th edition.

Question

I was reading the solution of this question:

Show that $L^2[a,b]$ is of the first category

But I do not know: What theorem said that $L^2[a,b]$ is dense in $L^1[a,b]$ in Royden "Real analysis " 4th edition. Could anyone help me in finding this theorem ?

Also, could anyone tell me the general idea of the solution given in this link please?

The proof of that statement is described in the answer you linked. For $f\in L^1[a,b]$ check that $f\cdot \Bbb1_{|f|^{-1}([0,n]}$ is in $L^2$ and that this converges to $f$ in $L^1$ sense. — s.harp, Mar 18 '20 at 12:18

score 1 · Accepted Answer · answered Mar 18 '20 at 12:19

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If $f \in L^{1}[a,b]$ then $f_n =fI_{\{|f| \leq n}\}$ is sequence in $L^{2}[a,b]$ converging in $L^{1}[a,b]$ to $f$.

answered Mar 18 '20 at 12:19

Kavi Rama Murthy

359,332

Why we are sure of the existence of this sequence $f_{n}$ and why we are sure that it is in $L^2$? – Mar 18 '20 at 12:24
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@Mathstupid $\int|f_n|^{2} \leq n\int |f|$ so $f_n \in L^{2}$ for each $n$. $f_n \to f$ in $L^{1}$ by DCT. – Kavi Rama Murthy Mar 18 '20 at 12:25
Do we have to use the open mapping theorem in the proof? – Mar 18 '20 at 12:35
Doesn't this also show the stronger fact that $L^\infty([a,b])$ is dense in $L^1([a,b])$ – Maximilian Janisch Mar 18 '20 at 13:31
Yes it does indeed. – Kavi Rama Murthy Mar 18 '20 at 13:45

What theorem said that $L^2[a,b]$ is dense in $L^1[a,b]$ in Royden "Real analysis " 4th edition.

1 Answers1