# Show that $f(x):=\sum\limits_{n=0}^{\infty}\frac{1}{n}h(2^{n}x),$ where $h$ is a piecewise function, converges uniformly on $[0,1]$

For $$x\in\mathbb{R}$$, consider a piecewise function defined by $$h(x):=\left\{ \begin{array}{ll} x,\ \ \ 0\leq x\leq 1\\ 2-x,\ \ 1\leq x\leq 2\\ 0,\ \ \text{otherwise}. \end{array} \right.$$

Now, consider the series $$f(x):=\sum_{n=0}^{\infty}\dfrac{1}{n}h(2^{n}x).$$ I'd like to show that the series converges uniformly on $$[0,1]$$.

I have an attempt to use Weierstrass M-test but failed. Here is how I tried:

Note that $$h(2^{n}x)=2^{n}x\mathbb{1}_{[0,2^{-n}]}(x)+(2-2^{n}x)\mathbb{1}_{[2^{-n},2^{-n+1}]}(x).$$ Therefore, for all $$x\in [0,1]$$ and for each $$n$$, we have \begin{align*} \Big|\dfrac{1}{n}h(2^{n}x)\Big|&\leq \dfrac{2^{n}}{n}|x|\mathbb{1}_{[0,2^{-n}]}(x)+\dfrac{2^{n}+2^{n}|x|}{n}\mathbb{1}_{[2^{-n},2^{-n+1}]}(x)\\ &\leq \dfrac{2^{n}\cdot 2^{-n}}{n}\mathbb{1}_{[0,2^{-n}]}+\dfrac{2+2^{n}\cdot 2^{-n+1}}{2}\mathbb{1}_{[2^{-n},2^{-n+1}]}(x)\\ &=\dfrac{1}{n}\mathbb{1}_{[0,2^{-n}]}(x)+\dfrac{4}{n}\mathbb{1}_{(2^{-n},2^{-n+1})}(x)\\ &\leq \dfrac{5}{n}. \end{align*}

My idea was to use $$\frac{5}{n}$$ as $$M_{n}$$, but the problem is that then $$\sum_{n=1}^{\infty}M_{n}$$ is divergent since it is a harmonic series.

It seems that the $$\frac{1}{n}$$ is really irritating. There is any other way to have a $$\frac{1}{n^{2}}$$? or I am heading to a wrong direction?

Thank you!

(Side remark: Having a $$\frac1n$$ term and starting the summation at $$n=0$$ is not a good idea. I assume the summations starts at $$n=1$$).

For $$x \le 0$$ and $$x \ge 1$$ we have $$h(2^nx) = 0$$ for all $$n \ge 1$$, so the series is just summing zeros for those $$x$$.

For $$0 < x < 1$$ let $$n_x$$ be the largest nonnegative integer $$n$$ with $$2^nx \le 1$$. We get

$$\sum_{n=1}^{n_x}h(2^nx)=\sum_{i=0}^{n_x-1}h(2^{(n_x-i)}x) \le \sum_{i=0}^{\infty}h(2^{(n_x-i)}x) = \sum_{i=0}^{\infty}2^{(n_x-i)}x = \sum_{i=0}^{\infty}\left(\frac12\right)^i2^{n_x}x = 2\times2^{n_x}x \le 2.$$

We start with rearranging the finite sum to go "from end to start", then add more nonnegative values and make it an infite sum. Since all arguments $$a$$ of $$h$$ in that sum are in the interval $$(0,1)$$, we have $$h(a)=a$$, which gives a standard infinte geometric series.

Note that by definition of $$n_x$$ we have that $$1 < 2^{n_x+1}x \le 2$$, so $$h(2^{n_x+1}x) \le 1$$ and finally $$2^{n_x+2}x > 2$$, so $$\forall k \ge 2: h(2^{n_x+k}x) = 0$$.

Alltogether, that means

$$\forall x \in \mathbb R:\sum_{n=1}^{\infty}h(2^nx) \le 3 \tag{1} \label{eq1}.$$

Now, to prove uniform convergence of $$\sum_{n=1}^{\infty}\frac1nh(2^nx)$$, we need to show that given any $$\epsilon > 0$$, we can find an $$N \in \mathbb N$$ such that

$$\forall x \in \mathbb R:\sum_{n=N}^{\infty}\frac1nh(2^nx) < \epsilon.$$

But using \eqref{eq1}, that's easy, just choose $$N > \frac3\epsilon$$:

$$\forall x \in \mathbb R:\sum_{n=N}^{\infty}\frac1nh(2^nx) \le \sum_{n=N}^{\infty}\frac1Nh(2^nx) =\frac1N\sum_{n=N}^{\infty}h(2^nx) \le \frac1N\sum_{n=1}^{\infty}h(2^nx) \le \frac3N <\epsilon.$$

This concludes the proof.

• sorry for the late reply. Thanks for the great answer!! How did you come up with this idea?? This is amazing. Jun 21 '20 at 14:17
• First is the observation that the M-test isn't likely to be of use as the best one can come up with is that $M_n \le \frac1n$, which leads to a diverging series. The next observation is that for $x$ with $0 < x \le 1$ the values $2^nx$ will at first also be in that interval, then one will be in $(1,2]$ and the remaining (infinitely) ones in $(2,\infty)$, which have function value 0. The final idea and that one that I took most time to come up with was to "turn around" the part of the sum until $n_x$ and realize that this creates a geometric series that is easily bounded. Jun 21 '20 at 15:26
• Brilliant!! Thank you! Jun 21 '20 at 17:31