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In a previous problem, I showed that $f_n\left(x\right)=n\left(\sqrt[n]{x}-1\right)$ converges pointwise to $\ln x$ on $\left(0,\infty\right)$ and uniformly on $\left[e^{-A},e^A\right]$. I am now trying to show that its inverse, $f_n^{-1}\left(x\right)=\left(x/n+1\right)^n$, converges pointwise to $f^{-1}\left(x\right)=e^x$ on $\mathbb{R}$ and uniformly on $\left[-A,A\right]$.

Now, what I have in mind is the following: $\lim_{n\to\infty}\left(\frac{x}{n}+1\right)^n=\exp\left(\lim_{n\to\infty}n\ln\left(\frac{x}{n}+1\right)\right).$

If we let $t=1/n$, then we get $\exp\left(\lim_{t\to0}\frac{1}{t}\ln\left(tx+1\right)\right)=\exp\left(\frac{0}{0}\right).$

Therefore, we can apply L'Hôpital's rule: $\exp\left(\lim_{t\to0}\frac{x}{tx+1}\right)=e^x.$

Is this correct? I have a feeling that, in the first step, I may not exponentiate a limit like that.

Furthermore, when it comes to showing uniform convergence, I know that I need to prove that for every $\epsilon>0$, there is an $N\in\mathbb{N}$ such that $\left|f_n^{-1}(x)-f^{-1}(x)\right|=\left|\left(\frac{x}{n}+1\right)^n-e^x\right|<\epsilon$whenever $n\geq N$ and $x\in[-A,A]$.

However, in that last problem, I had to use an esoteric definition of the exponential, and I am afraid that I may need to do the same for this problem. Do you guys have any ideas? Thanks a whole lot!

By the way, the book hints that I use the MVT, but I am clueless as to how to apply it here.

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    Josue: I am sorry if my words had that affect on you. In general, you should never feel "dumb" if there is something you just "don't see." Mathematics is full of things that become obvious only AFTER you spend a lot of time working through not-so-obvious details. I was assuming that you were familiar with Arzela-Ascoli (if you have time, do follow the link I gave above to wikipedia article about it). It is possible to solve your problem without applying Arzela-Ascoli directly; see my answer below.2012-03-14

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Above I suggested the use of Arzela-Ascoli theorem, which the original poster said they hadn't covered. Let us carry through without the use of Arzela-Ascoli, though as anyone familiar with Arzela-Ascoli will notice that in this special case our approach below simply "hides" Arzela-Ascoli behind some rather elementary arguments.

So, with $f_n(x) = \left(1 + x/n\right)$^n, we wish to show that the sequence $\{f_n(\cdot)\}_{n\in\mathbb{N}}$ converges to $\exp(\cdot)$ on any compact (i.e. closed and bounded) interval $[a, b]$.

Let $\epsilon > 0$. Fix $x_0\in [a, b]$ and let $N_0\in\mathbb{N}$ be such that for all $n\geq N_0$, $|f_n(x_0) - e^{x_0}| < \epsilon/2$. Consider, for any $x, y \in [a, b]$:

$ |(f_n(x) - e^x) - (f_n(y) - e^y)|\leq |f_n(x) - f_n(y)| + |e^x - e^y|.$

By the MVT, the right had side of the above inequality is

|f'_n(z_n)||x - y| + e^{w_n}|x - y|,

where $z_n, w_n\in (x, y)$, assuming $x < y$. On the other hand, for any $n$,

f'_n(x) = \left(\frac{x}{n} + 1\right)^{n-1}.

Observe that for all $x\in [a, b]$, f'_n(x) is uniformly bounded in $n$; that is, there exists $C_0 > 0$ such that for all $x\in [a, b]$ and $n\in\mathbb{N}$, |f'_n(x)|< C_0 (prove this to convince yourself of this fact). Also, since $[a, b]$ is a bounded interval, $e^x\leq e^b$ for all $x\in[a, b]$. Hence we get:

|f'_n(z_n)||x - y| + e^{w_n}|x - y|\leq C|x - y|,

where $C = \max\{C_0, e^b\}$.

Now, if $y \in (x_0 - \epsilon/2C, x_0 + \epsilon/2C)$, then from above we obtain:

$|(f_n(x_0) - e^{x_0}) - (f_n(y) - e^y)| < \epsilon/2.$

Therefore,

$|f_n(y) - e^y| \leq |f_n(x_0) - e^{x_0}| + \epsilon/2 < \epsilon.$

In effect, we have proved the following:

Lemma: For any $\epsilon > 0$ and any $x_0\in [a, b]$, there exists $\delta = \delta(x_0) > 0$ and $N_0\in\mathbb{N}$, such that for all $n\geq N_0$ and $y\in (x_0 - \delta, x_0 + \delta)$,

$|f_n(y) - e^y| < \epsilon.$

Now, fix $\epsilon > 0$. Apply the Lemma above to every $x\in [a, b]$. Then for each $x\in[a, b]$ we get a nonempty open interval $I(x)$ centered at $x$, and $N_x \in \mathbb{N}$, such that for all $y\in I(x)$ and $n\geq N_x$, $|f_n(y) - e^y| < \epsilon$. On the other hand, $[a, b]$ is compact; in other words, you need only finitely many $x_0, x_1, \dots, x_m\in[a,b]$, such that $[a, b]\subset \cup_{i=1}^m I(x_i)$. Now take $N = \max_{1\leq i\leq m}\{N_{x_i}\}$. Then for any $y\in[a,b]$ and $n\geq N$, we get $|f_n(y) - e^y| < \epsilon$.

This proves uniform convergence of the sequence $\{f_n(\cdot)\}$ to $\exp(\cdot)$ on any compact interval $[a, b]$. We're done.

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Some hints:

  • Show that for $t>-1$ we have $t-\frac{t^2}2\leq \ln(1+t)\leq t$.
  • Use this to deduce that for $x\in [-A,A]$ and $n\geq A+1$: $\frac xn-\frac{x^2}{2n^2}\leq \ln\left(1+\frac xn\right)\leq\frac xn.$
  • Find a bound for $\sup_{x\in [-A,A]}|\left(1+\frac xn\right)^n-e^x|$ which will allow us to conclude.
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    @JosuéMolina In fact I meant logarithm in base $e$, so I should write $\ln$. Thanks!2012-03-14