Exercise involving pointwise and uniform convergence: why does it converge and why it does not converge?

Question

(a) Let $(f_{n})_{n=1}^{\infty}$ be a sequence of functions from one metric space $(X,d_{X})$ to another $(Y,d_{Y})$, and let $f:X\to Y$ be another function from $X$ to $Y$. Show that if $f_{n}$ converges uniformly to $f$, then $f_{n}$ also converges pointwise to $f$.

(b) For each integer $n\geq 1$, let $f_{n}:(-1,1)\to\textbf{R}$ be the function $f_{n}(x) = x^{n}$. Prove that $f_{n}$ converges point wise to the zero function, but does not converge uniformly to any function $f:(-1,1)\to\textbf{R}$.

(c) Let $g:(-1,1)\to\textbf{R}$ be the function $g(x) = x/(1-x)$. With the notation as in $(b)$, show that the partial sums $\sum_{n=1}^{N}f_{n}$ converges point-wise as $N\to+\infty$ to $g$, but does not converge uniformly to $g$ on the open interval $(-1,1)$.

MY ATTEMPT

(a) According to the definition of uniform convergence, for $\varepsilon > 0$ there is a natural number $N\geq 1$ such that for every $x\in X$ we have that \begin{align*} n\geq N \Rightarrow d_{Y}(f_{n}(x),f(x)) < \varepsilon \end{align*}

Since it holds for every $x\in X$, it converges point wise for each $x_{0}\in X$ separately.

(b) Let us tackle the problem in three parts: $x\in(-1,0)$, $x = 0$ and $x\in(0,1)$.

When $x\in(0,1)$, we have that \begin{align*} 0 < x < 1 \Rightarrow 0 < x^{2} < x < 1 \Rightarrow 0 < x^{3} < x^{2} < x < 1 \Rightarrow \ldots \end{align*} That is to say, $x^{n}$ is decreasing and bounded below by $0$. Consequently, it converges to some real number $L$. More precisely, we have that \begin{align*} L = \lim_{n\rightarrow\infty}x^{n+1} = \lim_{n\rightarrow\infty} x\times x^{n} = x\times\lim_{n\rightarrow }x^{n} = xL \Longleftrightarrow L(1 - x) = 0 \end{align*} Given that $x\in(0,1)$, we conclude that $L = 0$, and we are done.

On the other hand, if $x = 0$, then $f_{n}(x) = 0$. Hence $x^{n}$ converges to $0$.

Finally, we must consider $x\in(-1,0)$, that is to say, $-x\in(0,1)$. Since the series \begin{align*} \sum_{n=1}^{\infty}x^{n} = \sum_{n=1}^{\infty}(-1)^{n}(-x)^{n} \end{align*} converges according to the Leibniz test, we conclude that $x^{n}$ converges to 0.

Gathering all the previous results, we conclude that $f_{n}$ converges point-wise to the zero function on $(-1,1)$.

However I am not able to prove that it does not converge uniformly to any function $f$ defined on $(-1,1)$.

Can someone help me with this?

(c) Once again, we have that \begin{align*} \sum_{n=1}^{N}f_{n}(x) = x + x^{2} + \ldots + x^{N} = \frac{x(1 - x^{N})}{1-x} \end{align*} which is well defined for every $x\in(-1,1)$. Since $x^{N}\to 0$ when $x\in(-1,1)$, $\sum f_{n}\to g$, and the desired result follows.

Once again, I am not able to prove that it does not converges uniformly to $g$ on $(-1,1)$.

Can someone help me with this?

Any comments our alternative solutions are welcome as well.

Answer 1

To show $f_n(x), x \in E$ do not uniformly converge to $f(x)$, the standard trick is to show $\sup_{x \in E} |f_n(x) - f(x)|$ fails to converge to $0$ as $n \to \infty$.

With this in mind, to show $x^n$ do not uniformly converge to $0$, we can proceed as follows: \begin{align} \sup_{|x| < 1} |x^n| \geq \left(1 - \frac{1}{n}\right)^n \to e^{-1} \neq 0 \text{ as } n \to \infty. \quad (\text{since } |1 - n^{-1}| < 1) \end{align}

Using the same trick, I believe you can complete part (c) easily.

Answer 2

For the remainder of part b), choose some $\varepsilon \in (0, 1)$ (you can choose any such number). If $x^n$ converges uniformly to $0$, then we expect there to be some $n$ such that $x^n \in (0 - \varepsilon, 0 + \varepsilon)$ for all $x \in (-1, 1)$. (Indeed, we actually expect this to be true for all sufficiently large $n$, but we only need one value of $n$ for this argument.)

The problem is, as you head closer to $1$, it takes longer and longer for $x^n$ to lie in this interval. In particular, if we consider: $$x_0 = \varepsilon^{\frac{1}{n+1}} \in (0, 1),$$ then $x_0^n = \varepsilon^{\frac{n}{n+1}} > \varepsilon$, which contradicts our choice of $n$. So, there can never be even a single $n$ so that $x^n$ is uniformly closer than $\varepsilon$ to $0$, which strongly contradicts uniform convergence.

For the remainder of part c), suppose we have uniform convergence. We will use this supposition to show that $x^n \to 0$ uniformly, contradicting part b).

Fix $\varepsilon > 0$. Then we have an $N$ such that, for all $x \in (-1, 1)$, $$n \ge N \implies \left|\sum_{i=1}^n x^i - \frac{x}{1 - x}\right| < \frac{\varepsilon}{2}.$$ Choose any $n \ge N + 1$. Then $n - 1 \ge N$, hence for all $x \in (-1, 1)$, \begin{align*} \varepsilon &= \frac{\varepsilon}{2} + \frac{\varepsilon}{2} \\ &> \left|\sum_{i=1}^n x^i - \frac{x}{1 - x}\right| + \left|\sum_{i=1}^{n - 1} x^i - \frac{x}{1 - x}\right| \\ &\ge \left|\left(\sum_{i=1}^n x^i - \frac{x}{1 - x}\right) - \left(\sum_{i=1}^{n-1} x^i - \frac{x}{1 - x}\right)\right| \\ &= |x^n| = |x^n - 0|. \end{align*} So, $n \ge N + 1 \implies |x^n - 0| < \varepsilon$, i.e. $x_n \to 0$ uniformly, contradicting b).