1
$\begingroup$

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

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

and

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.

$\endgroup$
  • $\begingroup$ 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
2
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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
  • 1
    $\begingroup$ @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
  • 1
    $\begingroup$ @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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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