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Let $f_n(x)=e^{-(x-n)^2}$ and let $ g(x) = \begin{cases} \frac{1-e^{-x^2}}{x^2} & x \ne0 \\ 1 & x=0 \end{cases}$

Suppose $g$ is continious, bounded and have maximum at 0

Show that $\sum_{n=0}^{\infty}g\cdot f_n(x)$ converges uniformly.

I've managed to show it for $x=0$ by using geometric sums and Weierstrass M-test, but I can't really find a way to show it for $x\neq 0$

zhw.
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njlieta
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  • For $x=0$ wouldn't $\sum_{n=0}^{\infty}g\cdot f_n(0)=e^{-n^2}\rightarrow0$ as $n\rightarrow\infty$ or am I wrong here? – njlieta May 25 '16 at 20:11
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    I think the point is that it's strange (and kind of silly) to talk about uniform convergence at a single point. – zhw. May 25 '16 at 20:16
  • Oh yeah thats probably right, I guess a sum that convergence in a single point, would converge uniformly in such a point (or how to put it). I was just thinking the approach would be to show uniform convergence for $x=0$, $x<0$ and $x>0$ – njlieta May 25 '16 at 20:21
  • Already asked and already answered: http://math.stackexchange.com/questions/1799542/how-to-prove-the-inequality-frac1-e-x2x2e-x-n2-frac2n2/1799585#1799585 – Jack D'Aurizio May 25 '16 at 21:00

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