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Define $f:\mathbb{R}\rightarrow\mathbb{R}$ by $$f(x)=-2\sum_{n=1}^\infty xe^{-n^2x^2}$$

Is $f$ continuous at the origin? I think that $f$ is not continuous at the origin. Any help is appreciated.

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4 Answers 4

up vote 6 down vote accepted

For $N\geq1$ put $x_N:=-{1\over N}$. Then $$f(x_N)>2\sum_{n=1}^N {1\over N} e^{-(n/N)^2}=2\int_0^1 e^{-x^2}\ dx +o(1)\qquad(N\to\infty)\ .$$ It follows that $\liminf_{N\to\infty} f(x_N)=c$ for some $c>0$, whereas $f(0)=0$. Therefore $f$ is not continuous at $0$.

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Hint:
The function $$g_n(x)={ \dfrac{x}{e^{n^2x^2}}} $$ has local maxima at points $x_n$ such that $|x_n| =\dfrac{1}{\sqrt{2}n}.$

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For $x\ne 0$, let $N=\left\lfloor \frac1{|x|}\right\rfloor $. Then for the summands with $n\le N$ we have $e^{-n^2x^2}\ge e^{-1}$, hence $$\begin{align}|f(x)|=2\left|\sum_{n=1}^\infty x e^{-n^2x^2}\right|&=2|x|\sum_{n=1}^\infty e^{-n^2x^2}\\&\ge 2 |x|\sum_{n=1}^Ne^{-n^2x^2}\\&\ge 2|x|Ne^{-1}\\&\ge 2|x|\left(\frac1{|x|}-1\right)e^{-1}\\&=2(1-|x|)e^{-1}\\&\ge e^{-1}&\text{if }|x|\le\frac12\end{align}$$ As $f(0)=0$, the function $f$ is not continuous at $0$.

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For $\frac1{2N}<|x|<\frac1N$, $$ \begin{align} |f(x)| &=2\sum_{n=1}^\infty |x|\,e^{-n^2x^2}\\ &\ge2\sum_{n=1}^N |x|\,e^{-n^2x^2}\\ &\ge2\sum_{n=1}^N |x|\,e^{-1}\\ &=2N\,|x|\,e^{-1}\\ &\ge e^{-1}\tag{1} \end{align} $$ Suppose $f(x)$ is continuous at $x=0$. Then $$ \lim_{x\to0^+}f(x)=\lim_{x\to0^-}f(x)\tag{2} $$ Since $f(x)$ is odd, we have $$ \lim_{x\to0^+}f(x)=-\lim_{x\to0^-}f(x)\tag{3} $$ Thus, $(2)$ and $(3)$ imply that $$ \lim_{x\to0}f(x)=0\tag{4} $$ However, $(1)$ precludes $(4)$, so $f(x)$ can not be continuous at $x=0$.

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