- Let \(\psi ( x ) = x\) on \([ 0 , \frac { 1 } { 2 } ] \ , \ \psi ( x ) = 1 - x\) on \([ \textstyle { \frac { 1 } { 2 } } , 1 ]\) and extended periodically of period 1. Define \(\begin{array} { r } { f ( x ) = \sum _ { n = 0 } ^ { \infty } 2 ^ { - n } \psi ( 8 ^ { n } x ) } \end{array}\)
Show that \(f ( x )\) is continuous everywhere.
Show that \(f ( x )\) is differentiable nowhere.
Hint: Consider the difference quotients
\[ \Delta _ { h } f ( x ) \equiv { \frac { f ( x + h ) - f ( x ) } { h } } \]
where \(h = \pm 8 ^ { - k }\) and the sign is chosen so that x and \(x + h\) lie on the same linear segment of the graph of \(\psi ( 8 ^ { k - 1 } x )\) . Then
\(\begin{array} { r } { \Delta _ { h } f ( x ) = \sum _ { n = 0 } ^ { k - 1 } 2 ^ { - n } \Delta _ { h } \psi ( 8 ^ { n } x ) } \end{array}\)
\(\begin{array} { r } { | \Delta _ { h } f ( x ) | \geq 4 ^ { k - 1 } - \sum _ { n = 0 } ^ { k - 2 } 4 ^ { n } } \end{array}\)
Solution
1. The series converges uniformly, hence defines a continuous function.
The periodic triangle wave satisfies \[ 0\le \psi(x)\le \frac12 \] for every \(x\). Hence \[ \left|2^{-n}\psi(8^n x)\right|\le 2^{-n-1} \] uniformly in \(x\). Since \[ \sum_{n=0}^\infty 2^{-n-1}<\infty, \] the Weierstrass \(M\)-test gives uniform convergence of \[ f(x)=\sum_{n=0}^\infty 2^{-n}\psi(8^n x). \] Each summand is continuous, so \(f\) is continuous on \(\mathbb R\).
2. Construct small increments with a large difference quotient.
Fix \(x\in\mathbb R\) and \(k\ge1\). Choose \[ h_k\in\{8^{-k},-8^{-k}\} \] so that \(8^{k-1}x\) and \(8^{k-1}(x+h_k)\) lie in one linear piece of the periodic triangle wave \(\psi\). This is always possible because the displacement in the argument is only \(1/8\), whereas every linear piece has length \(1/2\).
For \(n\ge k\), \[ 8^n h_k=\pm 8^{n-k}\in\mathbb Z, \] so periodicity gives \[ \psi(8^n(x+h_k))=\psi(8^n x). \] Thus \[ \Delta_{h_k}f(x) =\sum_{n=0}^{k-1}2^{-n}\Delta_{h_k}\bigl(\psi(8^n x)\bigr). \]
Because \(\psi\) is \(1\)-Lipschitz, \[ \left|\Delta_{h_k}\bigl(\psi(8^n x)\bigr)\right| \le 8^n \] for \(n<k-1\). On the other hand, by the choice of sign, the \(n=k-1\) term lies entirely in one linear piece of slope \(\pm1\), so \[ \left|2^{-(k-1)}\Delta_{h_k}\bigl(\psi(8^{k-1}x)\bigr)\right| =2^{-(k-1)}8^{k-1}=4^{k-1}. \] Therefore \[ \begin{aligned} |\Delta_{h_k}f(x)| &\ge 4^{k-1}-\sum_{n=0}^{k-2}2^{-n}8^n\\ &=4^{k-1}-\sum_{n=0}^{k-2}4^n\\ &=4^{k-1}-\frac{4^{k-1}-1}{3}\\ &=\frac{2\cdot4^{k-1}+1}{3}. \end{aligned} \] The right-hand side tends to \(\infty\) as \(k\to\infty\).
3. Conclude nowhere differentiability.
Since \(|h_k|=8^{-k}\to0\) while \[ |\Delta_{h_k}f(x)|\to\infty, \] the difference quotients at \(x\) cannot converge to a finite limit. Thus \(f\) is not differentiable at \(x\).
The point \(x\) was arbitrary, so \(f\) is nowhere differentiable.