The functions i'm trying to determine the convergence properties of are

a) $f_n(x)=xe^{-nx}$ on $[0,\infty)$


b) $f_n(x)=nxe^{-nx}$ on $[0,\infty)$

for (a), $f_n(x)=x/e^{nx}$, it looks to me like this function will converge pointwise to zero as $n \rightarrow \infty$ because $e^{nx}$ will go to infinity and $x$ will be fixed because we are determining the pointwise convergence. I do not know what the intervals of uniform convergence will be, I'd like some help with that actually.

for (b), $f_n(x)=xn/e^{nx}$, I believe this function will also converge pointwise to zero because the exponential decay will decay faster than $n$ will grow. I also need help determining the intervals of uniform convergence for this series.

  • $\begingroup$ I'd imagine the ratio test will show these two are indeed convergent for the reasons you state - the exponential denominator will accelerate far faster that the linear numerator, and the ratio test will make that blatantly obvious as a result. $\endgroup$ – Rhys Hughes Jun 16 '18 at 0:33
  • $\begingroup$ yeah, cool. Can you help me determine on what intervals, if any, the convergence will be uniform? $\endgroup$ – Math is hard Jun 16 '18 at 0:40

You are correct that both converge pointwise to zero.

For uniform convergence, let's consider (b). It is not uniformly convergent on $[0,\infty).$ Note that $f_n$ converges uniformly to $f$ on $S$ means $\sup_{x\in S}|f_n(x)-f(x)| \to 0.$ Here our limit is $f(x)=0$ so this means $\sup_{x\in S}|f_n(x)| \to 0.$ We can see that this doesn't hold for $S=[0,\infty)$ since $f_n(x)$ has a maximum of $1/e$ at $x=1/n,$ so $\sup_{x\in [0,\infty)}|f_n(x)| = 1/e$ for all $n$ and this does not converge to zero.

When you consider other intervals $[a,b)$ where $a>0,$ consider that location the maximum of $f_n$ is $1/n$ so will eventually (when $n$ gets large enough) go out of the interval, and the maximum of $f_n$ on $[a,b)$ will occur at $x=a.$ (It may help to sketch the function if this is not obvious.)

Hopefully once you understand (b), (a) will be easier.

  • $\begingroup$ For part (a), It seems that the maximum value of the function will also occur at $1/n$. When plugging this into the original function, it seems that for a given n, the function will have a maximum value of $1/ne$, which tends to zero as $n \rightarrow \infty$, and so the function does indeed converge uniformly on $[0,\infty)$ $\endgroup$ – Math is hard Jun 16 '18 at 13:09
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    $\begingroup$ @MichaelVaughan Yep that's right. (a) is uniformly convergent on any subinterval of $[0,\infty)$ and (b) is uniformly convergent on subinterval that does not contain zero. $\endgroup$ – spaceisdarkgreen Jun 16 '18 at 16:31
  • $\begingroup$ how is (b) uniformly convegent on a subinterval that does not contain zero? The series of functions converges pointwise to zero on all of $(0,\infty)$, so how could it uniformly converge to anything? $\endgroup$ – Math is hard Jun 17 '18 at 14:30
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    $\begingroup$ @MichaelVaughn apologies, I meant to also exclude intervals with an open lower endpoint at zero. It uniformly converges to zero on, say, $[1,\infty).$ $\endgroup$ – spaceisdarkgreen Jun 17 '18 at 18:24

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