Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
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 $0$.

share|cite|improve this answer

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

share|cite|improve this answer

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 $0$.

share|cite|improve this answer

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 $x=0$.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.