Show that the series $\sum_{n=1}^{\infty} (-1)^{n}u_{n}(x)$ converges uniformly on $[0,1]$, but the series $\sum_{n=1}^{\infty} u_{n}(x)$ does not.

Question

Let $u_{n}(x)$ be defined by $$u_{n}(x)=\begin{cases}x^{n+1}\ln x,& \text{ if }x\in(0,1],\\0, &\text{ if }x=0.\end{cases}$$ Show that the series $\sum_{n=1}^{\infty} (-1)^{n}u_{n}(x)$ converges uniformly on $[0,1]$, but the series $\sum_{n=1}^{\infty} u_{n}(x)$ does not.

Can anyone give me a proof or some hints for this problem, please?

Excuse me for my latex, I don't know how to express it better.

Answer 1

On $x \in (0,1)$, the series $\sum_{n=1}^{\infty} u_n(x)$ is given by

$$ \sum_{n=1}^{\infty} x^{n+1}\ln(x) = x\ln(x) \sum_{n=1}^{\infty} x^{n} = x \ln(x) \left(\frac{1}{1 - x} - 1 \right) = \frac{x^2 \ln(x)}{1 - x}$$

which shows that the series converges pointwise on $(0,1)$ to $f(x) := \frac{x^2 \ln(x)}{1 - x}$. However, if the series $\sum_{n=1}^{\infty} u_n(x)$ would have converged uniformly on $[0,1]$, the limit function would be left-continuous at $x = 1$ (since each of the summands is) and then we would have

$$ -1 = \lim_{x \to 1^{-}} f(x) = \lim_{x \to 1^{-}} \sum_{n=1}^{\infty} u_n(x) = \sum_{n=1}^{\infty} u_n(1) = \sum_{n=1}^{\infty} 0 = 0.$$

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On $x \in (0,1)$, the partial sums of the series $\sum_{n=1}^{\infty} (-1)^n u_n(x)$ have the form

$$ S_N(x) = \sum_{n=1}^{N} (-1)^n x^{n+1} \ln(x) = x \ln(x) \sum_{n=1}^{N} (-x)^n = x\ln(x) \left( \frac{(-x)^{N+1} - (-x)}{(-x) - 1} \right) \\ = x \ln(x) \frac{(-x)^{N+1} + x}{1 + x}. $$

In particular, this shows that the series converges pointwise on $(0,1)$ to $g(x) := \frac{x^2\ln(x)}{1 + x}$ and in fact, also on $[0,1]$. Now, try to estimate

$$ \left| S_N - g(x) \right| = -\frac{x^{N+2}}{1+x} \ln(x) $$

and show that this goes to $0$ uniformly for $x \in [0,1]$.

Answer 2

Hint. Note that for $x\in (0,1)$ $$\sum_{n=1}^{N} (-1)^n u_n(x)=-\frac{x^2(1-(-x)^{N})\ln x}{1+x},$$ and $$\sum_{n=1}^{N} u_n(x)=\frac{x^2(1-x^{N})\ln x}{1-x}.$$ What are the pointwise limits of the two partial sums?