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If we have a sequence of bounded functions $f_{n}$ converging almost everywhere to another bounded function $f$ in a finite measure space such that

$$ |f_{n}(t)| \leq c$$ for some constant $c$. Then

$$ \int |f_{n}(t) - f(t)|^{2} d\mu$$ goes to zero.

Since we are in a finite measure space, the constants are integrable and hence DCT applies. However, it only gives the result that

$$ \int f_{n}(t)d\mu \rightarrow \int f(t)d\mu$$ and $$ \int |f_{n}(t) - f(t)| d\mu \rightarrow 0$$. How do we prove the convergence in $L^{2}$ norm?

Here is what I tried:

$$ \int |f_{n}(t) - f(t)|^{2} d\mu = \int (f_{n}-f)(\overline{f_{n}}-\overline{f})$$ $$ = \int |f_{n}|^{2} - \int f\overline{f_{n}} - \int f_{n}\overline{f} + \int |f|^{2} $$ which goes to zero by DCT (as $f_{n}\overline{f} \rightarrow |f|^{2}$ and so on). Am I right in all this or am I missing something? I am apprehensive because I intuitively feel that pointwise convergence should not imply $L^{2}$ convergence.

Stefan Hansen
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Vishal Gupta
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1 Answers1

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You are right, but you can also argue more directly. Let $g_n(t) = |f_n(t) - f(t)|^2$. Then $g_n \to 0$ almost everywhere and $|g_n| \le 4c^2$ almost everywhere. As we are in a finite measure space, $$ \|f_n - f\|_{L^2}^2 = \int g_n \, d\mu \to 0. $$

martini
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  • So that means $L^{2}$ convergence is implied by pointwise convergence? Is it true in infinite measure spaces also? – Vishal Gupta May 31 '13 at 07:00
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    No, it isn't, unless we have an $L^2$-function $h$ bounding $|f_n - f|$ for each $n$. Then use $|f_n - f|^2 \le h^2$ and $h^2 \in L^1$ to argue as above. In general: No. Let for example $f_n = \chi_{[n,n+1]} \in L^2(\mathbb R)$. Then $f_n \to 0$ pointwise, but $|f_n|_{L^2} = 1$, all $n$. – martini May 31 '13 at 07:02
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    No. In case of an infinite measure space, you cannot apply DCT ... as constants are not integrable, therefore not sufficient as a bound for DCT. – martini May 31 '13 at 07:08