$$ f_n(x)= \begin{cases} \left(1-\frac xn\right)^n & 0\le x<n \\ 0 & x\ge n \end{cases} $$

I am able to show that $f_n(x)\to e^{-x}$ pointwise on $[0,+\infty)$, and I wonder if my justification for uniform convergence is valid:

By properties of logarithm, I transformed the sequence into

$$ \left(1-\frac xn\right)^n=\exp\left\{n\log\left(1-\frac xn\right)\right\} $$

Employing the fact that for sufficiently large $n$

$$ \left|n\log\left(1-\frac xn\right)+x\right|=\left|n\left[-\frac xn+\mathcal O\left(1\over n^2\right)\right]+x\right|=\mathcal O\left(\frac1n\right) $$

we have $n\log\left(1-\frac xn\right)\to-x$ uniformly, hence for each $\delta>0$ there exists an $N$ such that for all $n>N$:

$$ |\log f_n(x)+x|<\delta $$

By the uniform continuity of the exponential function, for all $\varepsilon>0,x>0,y>0$, we have $|e^{-x}-e^{-y}|<\varepsilon$ whenever $|x-y|<\delta$. As a result let $y$ be $-\log f_n(x)$, so for all $\varepsilon>0,x>0$ there exists an $N$ such that for all $n>N$

$$ |f_n(x)-e^{-x}|<\varepsilon $$

I wonder if this proof is correct or please point out my mistake. Thank you all!

  • $\begingroup$ In this line here, $\left|n\log\left(1-\frac xn\right)+x\right|=\left|n\left[-\frac xn+\mathcal O\left(1\over n^2\right)\right]\right|=\mathcal O\left(\frac1n\right)$, I'm not sure why you wrote $O\left(\frac1n\right)$ at the end. $\endgroup$ – Connor Harris Nov 10 '20 at 3:52
  • $\begingroup$ @ConnorHarris I edited the question, there should be an $+x$ in the absolute value sign. $\endgroup$ – TravorLZH Nov 10 '20 at 5:16

Your proof is not valid. Your estimates are valid only for fixed $x$ and there is no uniformity. In particular $n \log (1-\frac x n) $ does not tend to $-x$ uniformly: If $|n log (1-\frac x n) +x| <\epsilon $ for $n \geq n_0$ and for all $x <n$ you get a contradiction by letting $x \to n$.

Hint for a valid proof: Choose $T$ such that $x>T$ implies $e^{-x} <\epsilon$. Note that $0 \leq f_n(x)\leq e^{-x} <\epsilon$ for all $n$ if $x>T$. So it is enough to consider $\sup_{x \leq T} |f_n(x)-e^{-x}|$ and here your approach can be used.

  • $\begingroup$ Did you mean that all I need is to show $f_n(x)\to e^{-x}$ uniformly on $[0,T]$ $\endgroup$ – TravorLZH Nov 10 '20 at 7:59
  • $\begingroup$ Your estimates are uniformly valid on $[0,T]$ so you should have no difficulty in completing the proof. @TravorLiu $\endgroup$ – Kavi Rama Murthy Nov 10 '20 at 8:01
  • $\begingroup$ Okay, that helps a lot $\endgroup$ – TravorLZH Nov 10 '20 at 8:03
  • $\begingroup$ Is it that choosing $T$ satisfying some quantity being $<\varepsilon$ the common way to determine uniform convergence of some sequence of functions on $[0,+\infty)$? $\endgroup$ – TravorLZH Nov 10 '20 at 8:35
  • 1
    $\begingroup$ Not always. It works in this case. @TravorLiu $\endgroup$ – Kavi Rama Murthy Nov 10 '20 at 8:37

I think you need to look closer at what's going on in the Taylor series. You can clean up the argument that $f_n(x):=n\log\left(1-\frac{x}{n}\right) \to -x$ uniformly, but this only works if $x\leq M$ for some $M>0$. To see this:

By Taylor's theorem, we know $$ f_n(x/n)=n\underbrace{f_n(0)}_{=0}+\underbrace{f_n'(0)}_{=-1}x+\frac{f_n''(c)}{2}\left(\frac{x^2}{n}\right) $$ for some $c$ between $0$ and $x/n$. Hence, \begin{align} |f_n(x/n)+x|\leq \left|\frac{f_n''(c)}{2}\right|\left(\frac{|x|^2}{n}\right)\leq \left|\frac{f_n''(c)}{2}\right|\left(\frac{M^2}{n}\right). \end{align} An easy calculation shows $f_n'(x)=\frac{-n}{1-\frac{x}{n}},\, f_n''(x)=\frac{-1}{(1-\frac{x}{n})^2}$. Now if $x$ is not demanded to be bounded above by some $M>0$ independent of $n$, then clearly $f_n''(c)$ is unbounded. However, with this demand, we have $$ |f_n''(c)|\leq \frac{1}{(1-\frac{M}{n})^2}\leq C<\infty $$ From here, your argument about the exponential being uniformly continuous allows you to assert that $e^{f_n(x)} \to e^{-x}$ uniformly on $[0,M]$.

Finally, given $\epsilon>0$, pick $M$, such that $e^{-x}<\epsilon$ for all $x>M$. Then for sufficiently large $n$, it is obvious that we have uniform convergence.


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