$\sum_{n=1}^\infty 2^{-n} f_n$ has unbounded variation on every $[a,b] \subset [0,1]$ with $a < b$

Question

Consider $n\in \Bbb N$, and define a piecewise linear function $f_n: [0,1] \to \Bbb R$ so that the graph of $f_n$ contains the points $\{(k \cdot 2^{-n}, (-1)^k)\}_{0\le k\le 2^n}$. Define $f: =\sum_{n=1}^\infty 2^{-n} f_n$. Since $|f_n| \le 1$ and $\sum_{n\ge 1} 2^{-n} < \infty$, $f$ is continuous. How would you show that $f$ has unbounded variation on every closed subinterval $[a,b]$ of $[0,1]$ with $a < b$?

The points $\{k \cdot 2^{-n}\}_{0\le k\le 2^n}$ form a partition of $[0,1]$ for every $n\ge 1$, so it seems a natural choice for a partition of $[a,b]$ to consider would consist of $\{a,b\}$ and points from $\{k \cdot 2^{-n}\}_{0\le k\le 2^n}$ that fall in the interval $[a,b]$.

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P.S. This is what $f_2$ looks like:

The plots of $\{f_n\}$ will just have more "oscillations" as $n$ increases.

Edit: The current answer looks good. However, is there a way to avoid using differentiability a.e. of functions of bounded variation - perhaps directly computing the variation?

Answer 1

By definition, $f_n$ can be periodically extended to the whole $\mathbb R$, and then $f$ is a continuous function defined on $\mathbb R$ with period $1$. This extension is merely for simplicity of the proof below. We claim that

The function $f$ is not differentiable at any $x_0\in\mathbb R$.

From the above claim, we know that $f$ is of unbounded variation on any closed interval, since a continuous function of bounded variation is differentiable almost everywhere, see the last line of the introduction part of this wikipieda page.

We prove the claim using the following lemma.

Lemma. Let $f: \mathbb R\to\mathbb R$ be a continuous function and $x_0\in\mathbb R$. Assume that $f$ is differentiable at $x_0$, then for any non-decreasing sequence $\{u_n\}_{n\geq1}$ and any non-increasing sequence $\{v_n\}_{n\geq1}$ with $u_n\leq x_0\leq v_n$ satisying $v_n-u_n>0$ and $\lim_{n\to\infty}(v_n-u_n)=0$, there holds $$\lim_{n\to\infty}\frac{f(v_n)-f(u_n)}{v_n-u_n}=f'(x_0).$$

It is not hard to prove this lemma, so we left the proof to reader.

Now we show that $f$ defined in OP is not differentiable at any point. Let $x_0$ be any real number. We assume that $f$ is differentiable at $x_0$ and we look for a contradiction. For each $n\in\mathbb N_{\geq1}$, there exists a unique $m_n\in\mathbb Z$ such that $\frac{m_n}{2^n}\leq x_0< \frac{m_n+1}{2^n}$. We take $u_n=\frac{m_n}{2^n}, v_n=\frac{m_n+1}{2^n}$, then $u_n\leq x_0< v_n$, $\{u_n\}$ is non-decreasing, $\{v_n\}$ is non-increasing, and $\lim_{n\to\infty}(v_n-u_n)=0$. By definition, $u_n$ and $v_n$ lie in the same segment of $f_n$, so they lie in the same segment of $f_k$ for all $k\leq n$. Hence, for $1\leq k\leq n$ we have $f_k(v_n)-f_k(u_n)=\pm 2^{k+1}(v_n-u_n)$. For $k\geq n+1$, we have $f_k(u_n)=(-1)^{2^{k-n}m_n}=1=(-1)^{2^{k-n}(m_n+1)}=f_k(v_n)$. Therefore, $$\frac12\frac{f(v_n)-f(u_n)}{v_n-u_n}=\frac12 \sum_{k=1}^n\frac1{2^k}\frac{f_k(v_n)-f_k(u_n)}{v_n-u_n}=\sum_{k=1}^n(\pm1).$$ As a result, if $n$ is odd, then $\frac12\frac{f(v_n)-f(u_n)}{v_n-u_n}$ is odd; if $n$ is even, then $\frac12\frac{f(v_n)-f(u_n)}{v_n-u_n}$ is even. Thus the limit $$\lim_{n\to\infty}\frac12\frac{f(v_n)-f(u_n)}{v_n-u_n}$$ does not exist. This contradicts to our assumption of the differentiability of $f$ at $x_0$.