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Let $ f(x) = \sum_{n=1}^{\infty} \frac{1}{n^2}e^{-n^2x^2}$

I know that the serie of the derivatives $g(x) = \sum_{n=1}^{\infty} -2xe^{-n^2x^2}$ is uniformely convergent on every set $]-\infty;-b] \cup [b;+\infty[ , b>0 $.

This implies that $f$ is differentiable on $\mathbb{R} \setminus \{0\}$.

My question is, is it differentiable on $0$ ?

I tried to apply L'hopistal rule but I cannot figure out if $g(x)$ has a limit when $x \to 0^+$ or $x \to 0^{-}$. I also tried to compute the derivative at $0$ but I don't see how to carry the computations.

Many thx for any help

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  • $\begingroup$ I'm confused about the definition of $f(x)$. Is $n$ or $x$ summed from 1 to $\infty$? $\endgroup$ – PiE Apr 2 '17 at 10:15
  • $\begingroup$ it is summed over $n$ sorry i will edit it $\endgroup$ – incas Apr 2 '17 at 10:42
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The function$$ f(x)=\sum_{n=1}^{\infty }\frac{1}{n^{2}}e^{-x^{2}n^{2}} $$ has a maximum at $x=0$, since $e^{-x^{2}n^{2}}\leq e^{0}=1$, so if $f$ is differentiable at $x=0$, necessarily, $f^{\prime }(0)=0$. For $m\in \mathbb{N }$ we have$$ \frac{f(0)-f(\frac{1}{m})}{\frac{1}{m}-0}=m\sum_{n=1}^{\infty }% \frac{1}{n^{2}}(1-e^{-\frac{n^{2}}{m^{2}}})\geq m\sum_{n=1}^{m} \frac{1}{n^{2}}(1-e^{-\frac{n^{2}}{m^{2}}}) $$ Consider the function $g(t)=1-e^{-t}-\frac{t}{2}$. Note that $g(0)=0$ and $% g(1)=\frac{1}{2}-e^{-1}>0$. Moreover, $g^{\prime }(t)=e^{-t}-\frac{1}{2}>0$ for $t<\log 2$. Thus, $g>0$ for $t\in (0,1)$ and so $1-e^{-t}\geq \frac{t}{2} $ for $t\in \lbrack 0,1]$. Using this inequality with $t=\frac{n^{2}}{m^{2}} \leq 1$ we get that$$ m\sum_{n=1}^{m}\frac{1}{n^{2}}(1-e^{-\frac{n^{2}}{m^{2}}})\geq m\sum_{n=1}^{m}\frac{1}{n^{2}}\frac{1}{2}\frac{n^{2}}{m^{2}}=\frac{ 1}{2}. $$ This shows that$$ \frac{f(0)-f(\frac{1}{m})}{\frac{1}{m}-0}\geq \frac{1}{2}, $$ that is $\frac{f(\frac{1}{m})-f(0)}{\frac{1}{m}-0}\leq -\frac{1}{2}$. Hence, $$\liminf_{x\rightarrow 0^{+}}\frac{f(x)-f(0)}{x-0}\leq -\frac{1}{2}, $$ which implies that $f$ cannot be differentiable at $x=0$.

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