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Let $\{f_n\}$ be a sequence of functions in $L^2[0,1]$ such that $f_n(x)\rightarrow 0$ almost everywhere on $[0,1]$. If $|f_n(x)|\le 1$ for all $n$ and almost all $x\in [0,1]$.

I require direction in answering the following questions:

(a) $$\lim_{n\rightarrow \infty} \int_0^1 f_n(x)g(x)~dx =0$$ for all $g\in L^2[0,1]$.

(b) Will the conclusion still hold if the condition $|f_n(x)|\le 1$ is replaced by the condition that $||f_n||_{L^2} $ are uniformly bounded?

Thanks for your responses.

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Hints: Cauchy-Schwarz ( and Dominated Convergence Theorem ( – JavaMan Nov 30 '11 at 0:43
up vote 2 down vote accepted

For (a), given any $g\in L^2[0,1]$, since $|f_n(x)|\le 1$ a.e. on $[0,1]$ for all $n$, we have $$|f_n(x)g(x)|\le |g(x)| \mbox{ a.e. on $[0,1]$ for all }n.$$ Since $g\in L^2[0,1]$, by dominated convergence theorem DCT, we have $$\lim_{n\rightarrow \infty} \int_0^1 f_n(x)g(x)~dx =\int_0^1 \lim_{n\rightarrow \infty}\big(f_n(x)g(x)\big)~dx=0,$$ where the last equality follows from the assumption that $f_n(x)\rightarrow 0$ a.e. on $[0,1]$ as $n\rightarrow\infty$.

For (b), we can see that the conclusion does not hold if we just assume that $||f_n||_{L^2}$ is uniformly bounded by the following example: take $f_n$ to be $$f_n(x)=\left\{ \begin{array}{ll} n, & \hbox{if } x\in[0,\frac{1}{n}]; \\ 0, & \hbox{otherwise.} \end{array} \right.$$ It's clear that $f_n(x)\rightarrow 0$ a.e. on $[0,1]$ as $n\rightarrow\infty$, and $\|f_n\|_{L^2}=1$ for all $n$. Take $g(x)=x^{-\frac{1}{3}}$. Then $g\in L^2[0,1]$ because $\int_0^1g(x)^2dx=\int_0^1x^{-\frac{2}{3}}=3x^{\frac{1}{3}}\Big|^1_0=3$. However, $$\int_0^1f_n(x)g(x)dx=n\int_0^{\frac{1}{n}}x^{-\frac{1}{3}}dx=\frac{3}{2}nx^{\frac{2}{3}}\Big|^{\frac{1}{n}}_0=\frac{3}{2}n^{\frac{1}{3}}\rightarrow\infty$$ as $n\rightarrow\infty$.

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How did you get $\|f_n\|_{L^2}=1$. I'm getting $\sqrt{n}$. I'm using this definition: $$ \|f_n\|_{L^p}=\left(\int |f_n|^p\right)^{1/p}.$$ Am I doing something wrong? – Kuku Mar 23 '12 at 1:35

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