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$g(x)$ is the periodic function defined as $g(x) = |x|$ for $x \in [-1,1]$; $g(x+2) = g(x)$. Define $$f(x) = \sum_{n=0}^{\infty} \frac{g(3^{n} x)}{3^{n}}$$

  1. Prove $\forall x \in \mathbb{R}$, $f(x)$ is uniformly continuous.
  2. Prove $\forall x \in \mathbb{R}$, $f(x)$ is not differentiable.

I really have no idea where to start. I am thinking that probably I need to find a closed form for $f(x)$ for the first problem, but I don't know how.

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    1. doesn't make sense. You want to say that $f$ is uniformly continuous on $\mathbb R.$ – 2017-02-17
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    @zhw. You are right, my fault. – 2017-02-17
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    @zhw. Would you mind give me some hints to start? – 2017-02-17

2 Answers 2

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  1. First observe that $|g(x)|\leq 1$ for all $x\in\Bbb{R}$. Thus by Weirestrass $M-$test $f(x)=\sum_{n=0}^\infty\frac{g(3^nx)}{3^n}$ is uniformly convergent. Moreover by continuity of $g$ we see that $f$ is continuous. To prove the uniform continuity, note that $f(x+2)=f(x)$, thus $f$ is periodic. A continuous periodic function is uniformly continuous, this proves the claim.

  2. Let $x\in\mathbb{R}$, note that the interval $(3^m(x-2\cdot3^{-m-2}),3^m(x+2\cdot3^{-m-2}))$ has length smaller than $1$ for every $m\in\mathbb{N}$, thus one of $(3^m(x-2\cdot3^{-m-2}),3^mx)$ and $(3^mx,3^m(x+2\cdot3^{-m-2}))$ contains no integer. Define the sequence $h_m$ such that $h_m=2\cdot3^{-m-2}$ if $(3^mx,3^m(x+2\cdot3^{-m-2}))$ contains no integer, and $h_m=-2\cdot3^{-m-2}$ otherwise. It's easy to see that $h_m\to 0$. Now let's look at $\frac{f(x+h_m)-f(x)}{h_m}$. We consider the following four cases:

    • If $n\geq m+2$, then by $g(x+2)=g(x)$ we have $$g(3^n(x+h_m))=g(3^nx\pm2\cdot3^{n-m-2})=g(3^nx),$$ therefore $$\frac{g(3^n(x+h_m))-g(3^nx)}{h_m}=0.$$

    • If $n=m+1$: we have $$g(3^{m-1}(x+h_m))-g(3^{m-1}x)=g(3^{m-1}x+\frac{2}{3})-g(3^{m-1}x)\leq\frac{2}{3}$$ where the last inequality follows from the simple observation: $$|g(t)-g(y)|\leq |t-y|,\quad\forall t,y\in\mathbb{R}\quad(*).$$

    • If $n=m$, then we know that there is no integer between $3^m(x+h_m)$ and $3^m x$. This means the two points $(3^m(x+h_m),g(3^m(x+h_m))$ and $(3^m x,g(3^m x))$ must lie on the same straight line segment on the graph of $g$, therefore $$\frac{|g(3^m(x+h_m)-g(3^m x)|}{h_m}=\frac{|3^m(x+h_m)-3^mx|}{|h_m|}=3^m.$$

    • If $n

Now put everything together: $$\begin{aligned}\frac{|f(x+h_m)-f(x)|}{|h_m|}&=\left|\sum_{n=0}^{m+1}\frac{g(3^n(x+h_m))-g(3^nx)}{h_m}\right|\\ &\geq -\frac{2}{3}+3^m-\sum_{n=0}^{m-1}3^n\\ &=-\frac{2}{3}+3^m-\frac{1-3^m}{1-3}\\ &=-\frac{1}{6}+\frac{3^m}{2}\\ &\to+\infty \end{aligned}$$ as $m\to 0$. Thus $f$ is not differentiable at $x$.

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    Thank you for telling me those important theorems. I actually did not learned those theorems. Your answer gives me a way that I never think about. – 2017-02-17
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I remember a problem in Rudin's PMA that is very similar to this one. The Theorem in the book is "there exists a real continuous function on the real line which is nowhere differentiable".

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Theorem 7.10 is just the weierstrauss m-test (https://en.wikipedia.org/wiki/Weierstrass_M-test) and theorem 7.12 just says that the limit of a uniformly convergent sequence of continuous functions is continuous. I know this problem is slightly different than yours, but they are so similar that I thought I'd post this

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    Thank you for telling me those important theorems. I am still in the chapter 6 of Rudin's PMA, so I actualy did not learned those theorems. – 2017-02-17
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    no problem. If you have the book, this is Theorem 7.18 of the third edition – 2017-02-17