Uniform Convergence of a Nonlinear Sequence of Functions #2

Question

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.

Answer 1

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.

Answer 2

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.