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Let $\mu$ be the Lebesgue measure on the segment $[0,1]$. Let $(f_n)$ be a sequence in $L^{2}([0,1])$ such that $f_n\rightarrow f$ p.w. and $f_{n+1}\geq f_{n}$. Does $f_n\rightarrow f$ also in the 2-norm? (what if $(f_n)$ is a sequence of simple functions?)

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I'm surprised that you know what $L^2$ is without knowing the monotone convergence theorem. Where did you learn this material from? By the way, what does "p.w." mean? – Jonas Teuwen Aug 21 '11 at 12:51
p.w. = pointwise I guess – Henri Aug 21 '11 at 12:56
But the pointwise limit of $L^2$ functions need not be $L^2.$ Consider the family given by $f_n(x) = \frac{1}{\sqrt{x+1/n}}.$ Then $f_n$ is a positive, nondecreasing sequence of $L^2([0,1])$ functions, but the pointwise limit ${1}/{\sqrt{x}}$ is not $L^2.$ Moreover, neither is $|f_n -f|$ for any $n.$ The monotone convergence theorem only answers the $L^1$ analog of mathfreaks question. – jspecter Aug 21 '11 at 13:09

This is what is called the Monotone Convergence Theorem, and it is one of the most important convergence theorems for integration of functions and summation of series.

In your particular case, $f_n\to f$ pointwise and monotonically increasing, so $f_n-f\to 0$ pointwise and monotonically increasing. This means that $|f_n-f|^2\to 0$ pointwise and monotonically decreasing. Thus, $\int_0^1|f_n(x)-f(x)|^2{\rm d}x\to 0$ by monotone convegence. If you want to stick with monotonically increasing functions, $-|f_n-f|^2\to 0$ pointwise and monotonically increasing.

I just noticed that nowhere was it specified that $\|f_n\|_{L^2}$ was bounded. If $\lim_{n\to\infty}\|f_n\|_{L^2}=\infty$, then $f\notin L^2$.

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Further explanation: $|f_n-f|^2$ decreases to zero pointwise. – GEdgar Aug 21 '11 at 13:21
Why? It seems to me that the monotone convergence theorem assures that $f_n \to f$ in the $1$-norm. Why also in $2$-norm? – Andrea Aug 21 '11 at 13:22
@GEdgar: sorry, I was working on that, but I was drawn away for a while. – robjohn Aug 21 '11 at 13:57
Monotone convergence requires both monotone increasingness and nonnegativity. – Zarrax Aug 21 '11 at 14:39
@Zarrax: I was using the version not requiring non-negativity, derived from the non-negative version: Let $f^+(x)=\left\{\begin{array}{cl}f(x)&\mathrm{if\ }f(x)\ge0\\0&\mathrm{if\ }f(x)<0\end{array}\right.$ and $f^-(x)=\left\{\begin{array}{cl}0&\mathrm{if\ }f(x)\ge0\\-f(x)&\mathrm{if\ }f(x)<0\end{array}\right.$ Then $\{f_n^+\}$ and $\{f_0^--f_n^-\}$ are non-negative, non-decreasing sequences of functions tending to $f^+$ and $f_0^--f^-$ respectively. Apply the non-negative restricted version to $f_n^++(f_0^--f_n^-)-f_0^-(=f_n)$ which converges to $f^++(f_0^--f^-)-f_0^-(=f)$ – robjohn Aug 22 '11 at 18:34

As long as the limit function $f$ is in $L^2$ this is true. Let $f_n^+$ and $f^+$ denote the positive parts of $f_n$ and $f$ respectively, and $f_n^-$ and $f^-$ the negative parts. Then by the monotonicity conditions you have $$|f_n| \leq |f_n^+| + |f_n^-| \leq |f^+| + |f_1^-|$$ $$|f| \leq |f^+| + |f^-| \leq |f^+| + |f_1^-|$$ So if $g(x)$ denotes $|f^+| + |f_1^-|$, then $g(x)$ is an $L^2$ function that dominates $|f|$ and each $|f_n|$. Hence $$|f_n - f|^2 \leq (|f_n| + |f|)^2 \leq 4g^2$$ Since $4g^2$ is in $L^1$, the dominated convergence theorem gives $$\lim_{n \rightarrow \infty} \int_0^1 |f_n - f|^2 = 0$$ This is the desired $L^2$ convergence.

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