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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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Since $\exp$ is continuous, you can exponentiate limits –  no identity Mar 13 '12 at 5:47
    
As far as uniform convergence: try using Arzela-Ascoli theorem (en.wikipedia.org/wiki/Arzel%C3%A0%E2%80%93Ascoli_theorem). In order to verify equicontinuity (so you can apply Arzela-Ascoli), you'll see that MVT comes quite naturally. Try it, and if you are still having trouble, let us know. –  William Mar 13 '12 at 7:07
    
I am totally confused... we have never been taught that theorem before, and my level of analysis is not that high to understand any suggestion I have been given thus far... I feel so foolish. –  Josué Molina Mar 13 '12 at 22:41
    
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. –  William Mar 14 '12 at 3:48

2 Answers 2

up vote 1 down vote accepted

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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Thank you very much, but I am confused: Where did $t-\frac{t^2}2\leq \ln(1+t)\leq t$ come from? –  Josué Molina Mar 13 '12 at 22:20
    
If I am not mistaken, I think it is an abstraction of the second step, but I still do not understand how you got to it. I have another question: Why are we taking the logarithm base ten? –  Josué Molina Mar 13 '12 at 23:08
    
I just noticed that $t-\frac{t^2}2\leq \ln(1+t)\leq t$ does not hold for $t\in\left(-1,0\right)$. –  Josué Molina Mar 14 '12 at 0:37
    
@JosuéMolina In fact I meant logarithm in base $e$, so I should write $\ln$. Thanks! –  Davide Giraudo Mar 14 '12 at 10:50

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