On a proof of the non-differentiability of the Blancmange curve

Question

In my nameless lecture notes, in the construction of a continuous, nowhere differentiable function (the Blancmange curve), we encounter the definition of the sequential derivative and other real-analysis concepts. I have some question about these concepts. Here's a brief background:

Part 1: the Blancmange curve

Let $$f_1(x)=\begin{cases} x\qquad \ \text{when } 0\leq x< 1/2,\\ 1-x \ \ \text{when } 1/2\leq x< 1. \end{cases}$$ and define $f_1(x)$ for other values of $x$ so that it becomes periodic with period $1$. Then we let $$f_2(x)=\frac12 f_1(2x).$$ This is a scaling of $f_1$; we get a function with half the period and half the amplitude. In general $$f_{k+1}(x)=2^{-k}f_1(2^kx)\quad \text{for all }x\in\mathbb{R},\quad k=1,2,\ldots.$$ Now sum up all these functions and let $$T(x)=\sum_{k=1}^\infty f_k(x).$$ Since $0\leq f_k(x)\leq 2^{-k}$, the convergence is uniform according to the Weierstrass $M$-test and hence $T(x)$ is continuous.

$T(x)$ (from Wikipedia)" />

To understand the next part, here's a lemma from Abbott's Understanding Analysis (second edition),

Lemma. Let $f$ be defined on an open interval $J$ and assume that $f$ is differentiable at some $a\in J$. If $(a_n)$ and $(b_n)$ are sequences satisfying $a_n<a<b_n$ and $\lim a_n = a = \lim b_n$, then $$f'(a) =\lim_{n\rightarrow\infty}\frac{f(b_n)-f(a_n)}{b_n-a_n}.$$

Part 2: the sequential derivative

We shall show that $T(x)$ is not differentiable at any arbitrary point $c$. For each integer $n$, we let $m$ be the integer that satisfies $$\frac{m}{2^n}\leq c <\frac{m+1}{2^n},\tag{1}$$ and denote by $I_n$ the interval $[m2^{-n},(m+1)2^{-n}]$. The length of this interval is $2^{-n}$ which tends to zero as $n\to\infty$, and the interval contains $c$. Hence if $T(x)$ is differentiable at $c$, then the difference quotient $$d_n=\frac{T((m+1)2^{-n})-T(m2^{-n})}{2^{-n}}$$ equals $T'(c)+o(2^{-n})$ which has the limit $T'(c)$ as $n\to\infty$.

1. Note that my lecture notes have a non-strict inequality, i.e. $\frac{m}{2^n}\color{red}{\leq}c<\frac{m+1}{2^n}$ and I think this is problematic in the proof of the lemma provided here, since we get division by zero if $a_n=a$, i.e. if $\frac{m}{2^n}=c$. On the other hand, consider $a=c=\frac14$ and $n=2$, then there is no integer $m$ satisfying $\frac{m}{4}<\frac14<\frac{m+1}{4}$. So it seems like we need the non-strict inequality after all ($m=1$ would do it if the first inequality was a non-strict inequality). This confuses me and I'd be grateful for a comment or two on this. Moreover I think $o(2^{-n})$ should be $o(1)$ (as suggested by this answer).

Part 3: proof

We shall prove that $d_n$ has no limit.

Consider the corresponding difference quotient for $f_k$. If $k>n$, then $f_k$ equals zero at both endpoints of $I_n$, and then the difference quotient equals $0$. This cannot happen if $k\leq n$, because then the difference quotient equals $1$ or $-1$, since $f_k$ is made up of line segments having these gradients. The sign depends only on the gradient of $f_k$ at the point $a$.

The difference quotient $d_n$ is equal to the sum of these numbers. The difference $d_n-d_{n-1}$ is a difference between two sums the terms of which are equal in pairs except the $n$th terms. Hence $$d_n-d_{n-1}=\pm 1.$$ This is valid for all $n$, and therefor the sequence $(d_n)$ has no limit.

2. By "...corresponding difference quotient for $f_k$" I assume they mean $$\frac{f_k((m+1)2^{-n})-f_k(m2^{-n})}{2^{-n}}.\tag{2}$$ Since $f_{k+1}(x)=2^{-k}f_1(2^kx)$, the above fraction simplifies to $$\frac{2^{-k+1}f_1(2^{k-1}(m+1)2^{-n})-2^{-k+1}f_1(2^{k-1}m2^{-n})}{2^{-n}}=2^{-p}\left(f_1(2^{p}(m+1))-f_1(2^{p}m)\right),$$ where $p=k-1-n$. When $k>n$, then $p\geq 0$ and then $2^{p}(m+1)$ and $2^{p}m$ are integers where $f_1$ is $0$. I have a hard time convincing myself of that $(2)$ equals $-1$ or $1$ when $k\leq n$. Then $p<0$, and $2^{p}(m+1)$ and $2^{p}m$ are rational numbers, but what is $f_1$ then?

3. Finally, and probably related to the previous question, I do not understand why $d_n-d_{n-1}=\pm 1$. Since $T(x)$ is uniformly convergent on $\mathbb{R}$ and the difference quotient of $f_k$ equals $0$ for $k>n$, we have $$\begin{align} d_n-d_{n-1}&=\frac{T((m+1)2^{-n})-T(m2^{-n})}{2^{-n}}-\frac{T((m+1)2^{-n+1})-T(m2^{-n+1})}{2^{-n+1}} \tag{3} \\ &= \frac{\sum_{k=1}^\infty f_k((m+1)2^{-n})-\sum_{k=1}^\infty f_k(m2^{-n})}{2^{-n}}-\frac{\sum_{k=1}^\infty f_k((m+1)2^{-n+1})-\sum_{k=1}^\infty f_k(m2^{-n+1})}{2^{-n+1}} \tag{4}\\ &=\sum_{k=1}^n \frac{ f_k((m+1)2^{-n})-f_k(m2^{-n})}{2^{-n}}-\sum_{k=1}^{n-1} \frac{ f_k((m+1)2^{-n+1})-f_k(m2^{-n+1})}{2^{-n+1}}, \tag{5} \end{align}$$ however, I do not see from this why the sums have equal terms up to the $n$th term.

Answer 1

1. The non-strict inequality is problematic

No. We can easily restate the lemma:

Suppose $J$ is an open set, $f:J\to\Bbb R$ is differentiable at the point $c\in J$, and suppose the sequences $(a_n)_{n\in\Bbb N},(b_n)_{n\in\Bbb N}\subset J$ both converge to $c$ and satisfy $a_n\le c<b_n$ for all $n\in\Bbb N$. Then: $$\lim_{n\to\infty}\frac{f(b_n)-f(a_n)}{b_n-a_n}=f'(c)$$

Why? Now, you're right that if $a_n=c$ for a particular $n$ then the method of the proof doesn't work due to division by zero. This is quite easily avoided. Because $f$ is differentiable at $c$, for any $\epsilon>0$ there is some $\delta>0$ such that $\frac{f(x)-f(c)}{x-c}-f'(c)\in(-\epsilon,\epsilon)$ whenever $x\in(c-\delta,c+\delta)$ and $x \neq c$. There is some $N\in\Bbb N$ with $a_n,b_n\in(c-\delta,c+\delta)$ for all $n\ge N$. So the linked proof shows that for all $n\ge N$:

Case $1$: $a_n < c$: $$\left|\frac{f(b_n)-f(a_n)}{b_n-a_n}-f'(c)\right|\le\left|\frac{f(b_n)-f(c)}{b_n-c}-f'(c)\right|+\left|\frac{f(c)-f(a_n)}{c-a_n}-f'(c)\right|<2\epsilon$$

Or:

Case $2$: $a_n=c$: $$\left|\frac{f(b_n)-f(a_n)}{b_n-a_n}-f'(c)\right|=\left|\frac{f(b_n)-f(c)}{b_n-c}-f'(c)\right|<\epsilon<2\epsilon$$

So, regardless of whether or not $a_n=c$, it is still true that the differences can be bounded by $2\epsilon$ for all $n\ge N$. So the stated limit is still correct.

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2. Why is: $$2^{-p}(f_1(2^p(m+1))-f_1(2^p m))\in\{-1,1\}$$When $p<0$?

Observe that at most one of $2^p(m+1),2^pm$ can be integral. In these cases, the above equals one of two values: $$2^{-p}(f_1(2^p(m+1))-0)=2^{-p}(f_1(2^p))=1\\2^{-p}(0-f_1(2^pm))=2^{-p}(-f_1(-2^p))=-1$$Using the defined periodicity of $f_1$.

Suppose neither are integral. Let $N=\lfloor 2^pm\rfloor$. It is necessarily the case that $N=\lfloor 2^p(m+1)\rfloor$ too (use the fact $m$ is integer).

We have two cases.

If $N<2^pm<2^p(m+1)\le N+1/2$, then we have: $$2^{-p}(f_1(2^p(m+1))-f_1(2^pm))=2^{-p}(2^p(m+1)-2^pm)=1$$

If $N+1/2\le 2^pm<2^p(m+1)<N+1$, then we have: $$2^{-p}(f_1(2^p(m+1))-f_1(2^pm))=2^{-p}(1-2^p(m+1)+2^pm-1)=-1$$

Overall we find the corresponding difference quotient to be $1$ if $N\le 2^pm<N+1/2$ and $-1$ otherwise. I get to rule out the possibility that $2^pm<N+1/2<2^p(m+1)$ as $m$ is integer.

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3. Why is $d_n-d_{n-1}=\pm1$?

I don't agree with your expansion, as it is misleading; the two "$m$"s involved are different, which I didn't spot the first time I answered this question. Let $m$ and $m'$ be the unique integers such that: $$\frac{m}{2^n}\le c<\frac{m+1}{2^n},\quad\frac{m'}{2^{n-1}}\le c<\frac{m'+1}{2^{n-1}}$$We can wonder about the relation between $m$ and $m'$. Clearly: $$\frac{2m'}{2^n}\le c<\frac{2m'+2}{2^n}$$So either $m=2m'$ or $m=2m'+1$ will hold.

Then with $p$ being a placeholder for $k-n$ we have: $$\begin{align}d_n-d_{n-1}&=\sum_{k=1}^\infty\frac{f_k((m+1)\cdot 2^{-n})-f_k(m\cdot 2^{-n})}{2^{-n}}\\&\quad-\sum_{k=1}^\infty\frac{f_k((m'+1)\cdot 2^{-(n-1)})-f_k(m'\cdot 2^{-(n-1)})}{2^{-(n-1)}}\\&=\sum_{k=1}^n\frac{f_1((m+1)\cdot 2^{p-1})-f_1(m\cdot 2^{p-1})}{2^{p-1}}\\&\quad-\sum_{k=1}^{n-1}\frac{f_1((m'+1)\cdot2^p)-f_1(m'\cdot2^p)}{2^p}\\&=2(f_1((m+1)/2)-f_1(m/2))\\&\quad+\sum_{k=1}^{n-1}\frac{2f_1((m+1)\cdot2^{p-1})-2f_1(m\cdot2^{p-1})-f_1((m'+1)\cdot2^p)+f_1(m'\cdot2^p)}{2^p}\end{align}$$If each and every summand in the final sum is zero, then $d_n-d_{n-1}$ would just equal the difference quotient $2(f_1((m+1)/2)-f_1(m/2))=\pm1$ ($p=-1$).

So, why are the summands zero? Let's work case by case. Assume $m=2m'$. Fix $1\le k\le n-1$, $p=k-n$. Write $N':=\lfloor2^pm'\rfloor$. Notice it suffices to show: $$2f_1((m'+1/2)\cdot2^p)-2f_1(m'\cdot2^p)\overset{?}{=}f_1((m'+1)\cdot2^p)-f_1(m'\cdot2^p)$$By previously used arguments, this is true and both sides equal $2^p$ in the instance that $N'\le 2^pm'<N'+1/2$; this is true and both sides equal $-2^p$ in the instance that $N'+1/2\le 2^pm'<N'+1$. What if $m=2m'+1$? Then we need to show: $$2f_1((m'+1)\cdot2^p)-2f_1((m'+1/2)\cdot2^p)\overset{?}{=}f_1((m'+1)\cdot2^p)-f_1(m'\cdot2^p)$$But this works just the same on the different cases $N'\le2^pm'<N'+1/2,N'+1/2\le 2^pm'<N'+1$. So all's well! The key point is that the step sizes of $2^p, (1/2)2^p$ are of the right size so that $(m'+1)\cdot2^p,(m'+1/2)\cdot2^p$ fall into the same "piece" of $f_1$ as $m'\cdot2^p$ always.

From this we can write: $$d_n=2\sum_{k=1}^n\left(f_1\left(\frac{m_k+1}{2}\right)-f_1\left(\frac{m_k}{2}\right)\right)$$Where $m_k:=\lfloor c\cdot2^k\rfloor$ for each $k$. This is evidently not Cauchy, each summand having value $1/2$ or $-1/2$. It is an interesting question whether $d_n$ actually diverges to an infinity, or just oscillates. I am not sure what the answer is.