Let $\phi:\mathbb R\mapsto\mathbb R$ is defined by $\phi(x)=\frac1{\sqrt{2\pi}}e^{-\frac{x^2}2}$ for all $x\in\mathbb R$.

Suppose we are given a sequence of functions $\{f_n\}$ such that $\lim_{x\to\pm\infty}f_n(x)\phi(x)=0$, $\{\int_{-\infty}^xwf_n(w)\phi(w)dw\}$ converges uniformly to $\int_{-\infty}^xwf(w)\phi(w)dw$ and $\{f_n\phi\}$ converges uniformly to $\{f\phi\}.$

I wish to have$$\int_{-\infty}^\infty x^2[f_n(x)-f(x)]^2\phi(x)dx$$ converges to $0$.

I have tried as follows:

Define $M_n=\sup_{x\in\mathbb R}\{f_n(x)\phi(x)-f(x)\phi(x)\}.$ Since $\{f_n\phi\}$ converges uniformly to $\{f\phi\}$ then $\{M_n\}\to 0.$ So the integral is bounded by $$\int_{-\infty}^\infty x^2M_n^2\frac1{\phi(x)}dx.$$ However, the above integral does not converges to $0$. Any suggestion or advice or correction?


Let for $n\geq 1$ $\displaystyle b_n=\sqrt{2\pi}\int_{-n}^n x^2\exp(x^2/2)dx$. Obviously, $b_n\to +\infty$ as $n\to \infty$. We can find a sequence $a_n>0$ such that $a_n\to 0$, and $a_n^2b_n\to +\infty$.

Now let $\displaystyle f_n(x)=\frac{a_n}{\phi(x)}$ for $|x|\leq n$, and $f_n(x)=0$ if $|x |>n$. If $|x|\to +\infty$, $f_n(x)\phi(x)\to 0$. With $f(x)=0$ we have $\displaystyle |f_n(x)\phi(x)-f(x)\phi(x)|\leq a_n$ for all $x$, and this show that $f_n\phi(x)\to f(x)\phi(x)$ uniformly on $\mathbb{R}$. But: $$\int_{-\infty}^{+\infty}x^2(f_n(x)-f(x))^2\phi(x)dx=a_n^2b_n\to +\infty$$

So If I am not mistaken, you must add some new hypothesis.

EDIT: I use your new hypothesis.

Put with the same notations as above $\displaystyle g_n(x)=\int_{-\infty}^x wf_n(w)\phi(w)dw$. We compute that $g_n(x)=0$ if $x\leq -n$, $\displaystyle g_n(x)=a_n\frac{x^2-n^2}{2}$ for $-n\leq x\leq n$, and $g_n(x)=0$ if $x\geq n$. As for $x\in [-n,n]$, we have $|x^2-n^2|\leq n^2$, to have $g_n(x)\to g(x)=0$ uniformly on $\mathbb{R}$, it suffice to have $a_n n^2\to 0$ (and this imply $a_n\to 0$). Hence we need on $a_n$ that $a_n n^2\to 0$ to verify your hypothesis. But $$ \int_{-n}^n x^2\exp(x^2/2)dx\geq \int_{1}^n x^2\exp(x^2/2)dx\geq \int_{1}^n x\exp(x^2/2)dx=\exp(n^2/2)-\exp(1/2)$$ and hence we can find such a sequence $a_n$ such that $a_n^2 b_n \to +\infty$. So if I am not wrong, the same counter-example seems to work.

  • $\begingroup$ Thank you very much @Kelenner for your time and help. However, I am very sorry, I missed a condition that I have added there. $\endgroup$ – Jlamprong Oct 3 '14 at 14:44

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