A friend of mine asked me the following question (which is motivated by an image processing problem of which I am unable to say more). Let $(f_n)_{n\geq 0}$ be an orthonormal sequence in $L^2([0,1])$, and define $F_n(x)=\int_0^x f_n(t)dt$. Is it true that $\frac{F_n}{\Vert F_n\Vert_2}\to 0$ weakly in $L^2$? Thanks in advance.
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2 Answers
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No, this isn't true even for $f_n(x) = \sqrt{2} \sin n\pi x$. Then $g_n := \frac{F_n}{\|F_n\|} = \sqrt{\frac{2}{3}}(1-\cos n\pi x)$ and $\langle g_n, 1\rangle = \sqrt{\frac{2}{3}} \not\to 0$.
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No.
Let $f_n$ take the value $2^{n/2+1}$ on $[1/4-2^{-n-2}, 1/4-2^{-n-3}]$, $-2^{n/2+1}$ on $[3/4+2^{-n-3}, 3/4+2^{-n-2}]$, and $0$ elsewhere. Then $F_n/||F_n||_2$ converges in $L^2$ to a function which takes the value $\sqrt{2}$ on $[1/4,3/4]$ and is $0$ elsewhere, and thus does not converge weakly to $0$.
answered Mar 23, 2013 at 21:49
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$\begingroup$ However, the question should be modified by replacing $F_n$ with $F_n-\int_0^1 F_n$. $\endgroup$– EtienneMar 24, 2013 at 17:14