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How can I prove that $$ \int_{\mathbb{R}^d} u(x) e^{-|x|^2} (e^{-|x|^2 / n} - 1)^2 dx \rightarrow 0 $$ as $n \rightarrow \infty$? Here $u \in L^2(\mathbb{R}^n)$.

I'm thinking the dominated convergence theorem, but I don't know how to bound the integrand.


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Good point. I've edited the question. I realize now that I made a mistake in the calculation that led to the original version. – maxpower Dec 23 '11 at 21:20
HINT Dominated convergence does work here. Did you try bounding the integrand by $u(x) e^{-|x|^2}$? Also, what is the pointwise limit of $u(x) e^{-|x|^2} (1 - e^{-|x|^2/n})^2$? – Srivatsan Dec 23 '11 at 21:25
up vote 3 down vote accepted

Added latter:(a simpler proof) We have $$\left(\exp\left(-\frac{|x|^2}n\right)\right)^2=\left(\int_0^{-\frac{|x|^2}n}e^tdt\right)^2\leq \frac{|x|^4}{n^2},$$ hence $$\left|\int_{\mathbb R^d}u(x)f_n(x)dx\right|\leq \frac 1{n^2}||u||_{L^2}\left(\int_{\mathbb R^d}e^{-2|x|^2}|x|^8dx\right)^{\frac 12},$$ and we can conclude since $e^{-2|x|^2}|x|^8$ is integrable. It's shows that the sequence $\{f_n\}$ converges strongly to $0$.

We can use dominated or the fact that the functions with compact support are dense in $L^2$. Put $f_n(x):=e^{-|x|^2}\left(\exp\left(-\frac{|x|^2}n\right)-1\right)^2$. Since $\left(\exp\left(-\frac{|x|^2}n\right)-1\right)^2\leq 1$ for all $x$ and for all $n$, we have $\lVert f_n\rVert_{L^2}\leq \sqrt{\int_{\mathbb R^d}e^{-2|x|^2}dx}=:M$. Let $u\in L^2(\mathbb R^d)$ and $\varepsilon >0$. Let $g$ continuous with compact support $K$ such that $\|u-g\|_{L^2}\leq\frac{\varepsilon}{M}$ (consequence of the monotone convergence theorem). Then \begin{align*} \left|\int_{\mathbb R^d}u(x)e^{-|x|^2}\left(\exp\left(-\frac{|x|^2}n\right)-1\right)^2dx\right|&\leq \int_{\mathbb R^d}|u-g|f_n(x)dx +\int_{\mathbb R^d}|g|f_n(x)\\ &\leq \|u-g\|_{L^2} M+\sup |g| \int_Ke^{-|x|^2}\frac{\sup_{w\in K} |w|^4}{n^2}dx\\ &\leq \varepsilon+\sup |g|\sup_{w\in K} |w|^4\frac 1{n^2}, \end{align*} so for all $\varepsilon >0$: $$\limsup_n\left|\int_{\mathbb R^d}u(x)e^{-|x|^2}\left(\exp\left(-\frac{|x|^2}n\right)-1\right)^2dx\right|\leq \varepsilon,$$ hence $$\lim_{n\to\infty}\left|\int_{\mathbb R^d}u(x)e^{-|x|^2}\left(\exp\left(-\frac{|x|^2}n\right)-1\right)^2dx\right|=0.$$

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+1. Let me point out that in fact, $(1 - \exp(|x|^2/n) )^2$ is bounded by $1$. :) Also, in your step $\lVert f_n\rVert_{L^2}\leq 4\int_{\mathbb R^d}e^{-2|x|^2}dx$, do you mean $\lVert f_n\rVert_{L^2}\leq 4\int_{\mathbb R^d}e^{-|x|^2}dx$? [I deleted my previous comment by the way.] – Srivatsan Dec 23 '11 at 22:12

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