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Yesterday I encountered the following exercise in a tutorial sheet from the University of Lyon : define a sequence of functions $(f_n)$ (with $f_n:[0,\infty) \to {\mathbb R}$) by $f_n(x)=\big(1-\frac{x}{n}\big)^n$ if $x< n$ and $0$ otherwise. It is easy and well-known that $(f_n)$ converges pointwise to $f(x)=e^{-x}$, and the exercise asks to show that the convergence is in fact uniform.

I found a solution as I explain below, but given the context I expect that a simpler solution exists, which is why I post this here on MSE.

My solution : as ${\sup}_{x\in[0,n]}|f_n(x)-f(x)|=e^{-n}$, it suffices to show that $M_n={\sup}_{x\in[0,n]}|f_n(x)-f(x)|$ tends to zero when $n\to+\infty$. We have $M_n=\sup_{y\in[0,1]} |g_n(y)|$ where $g_n(y)=e^{-ny}-(1-y)^n$. Using the inequality $e^{-y} \geq 1-y$ for $y\in[0,1]$, we see that $g_n\geq 0$ and we can remove the absolute value : $M_n=\sup_{y\in[0,1]} g_n(y)$. Next, $g'_n(y)=n((1-y)^{n-1}-e^{-ny})$, so that $g'_n(y)=0$ is equivalent to $(*)_n :h(y)=-\frac{n}{n-1}$ where $h(y)=\frac{\log(1-y)}{y}$. As $h'(y)=\frac{-\frac{y}{1-y}-\log(1-y)}{y^2}\leq 0$, we see that $(*)_n$ has a unique solution $y_n\in[0,1]$. A little computation yields $y_n=\frac{2}{n}-\frac{2}{3n^2}+o(\frac{1}{n^2})$, whence $M_n=g(y_n)=\frac{2e^{-2}}{3n}+o(\frac{1}{n})$. So $(M_n)\to 0$ as wished.

Related : Uniform Convergence of an Exponential Sequence of Functions

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Ewan Delanoy
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1 Answers1

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We have

$$\left(1- \frac{x}{n}\right)^n \leqslant e^{-x}\leqslant \left(1+ \frac{x}{n}\right)^{-n}.$$

Hence, using the Bernoulli inequality $(1 - x^2/n^2)^n \geqslant 1 - x^2/n,$

$$0 \leqslant e^{-x} - \left(1- \frac{x}{n}\right)^n = e^{-x}\left[1 - e^x\left(1- \frac{x}{n}\right)^{n}\right]\\ \leqslant e^{-x}\left[1 - \left(1+ \frac{x}{n}\right)^{n}\left(1- \frac{x}{n}\right)^{n}\right]\\= e^{-x}\left[1 - \left(1- \frac{x^2}{n^2}\right)^{n}\right]\leqslant e^{-x}\frac{x^2}{n}.$$

Therefore,

$$0 \leqslant\sup_{x \in [0,\infty)} \left|e^{-x} - \left(1- \frac{x}{n}\right)^n\right| \leqslant \sup_{x \in [0,\infty)}e^{-x}\frac{x^2}{n}= \frac{4e^{-2}}{n}\xrightarrow[n \rightarrow \infty]\quad 0.$$

The sequence $(f_n)$ is defined as

$$f_n(x) = \left(1- \frac{x}{n}\right)^n1_{x < n}.$$

Hence,

$$ 0 \leqslant \sup_{x \in [0,\infty)} \left|f_n(x) - e^{-x}\right| \leqslant \max\left(\frac{4e^{-2}}{n},1_{n < 2}\right)\xrightarrow[n \rightarrow \infty]\quad 0$$

RRL
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  • Your interval below the sup should be $[0,n]$ not $[0,\infty)$ – Ewan Delanoy Nov 23 '14 at 08:46
  • @EwanDelanoy: Thanks -- I just noticed your definition of $f_n$. I fixed above. The sup of $x^2e^{-x}$ is achieved at $x= 2$. So for $n \geq 2$ sup over $[0,\infty)$ is the same as sup over $[0,n]$ – RRL Nov 23 '14 at 09:10