Dini or Right or upper derivative of Weierstrass function Is it true that the right-side derivative of Weierstrass function, which is a classic example of continuous yet nowhere differentiable function, always non-negative? (In fact, positive)
That is, given $f(x) = \sum_{n=0}^\infty a^ncos(b^n\pi x)$ for $0<a<1$, and $b$ a positive odd with $ab>1+3\pi/2$, is it true that for every $x$,
$\lim\sup_{h\to0+} \frac{f(x+h)-f(x)}{h}\geq 0$?
Thanks.
 A: Your conjecture fails in two rather strong ways.
Corollary 11.4.1 on p. 128 of [1] states a result that is slightly stronger than the following, where $D^{+}f$ is the right upper Dini derivate of $f.$

Let $I$ be an open interval in $\mathbb R$ and let $f:I \rightarrow {\mathbb R}$ be a function. If $f$ is continuous and $D^{+}f \geq 0$ on $I$ (or even just $D^{+}f(x) \geq 0$ for all but countably many $x \in I),$ then $f$ is non-decreasing on $I.$ Hence, by Lebesgue's theorem on the differentiability of monotone functions, it follows that $f$ has a finite 2-sided derivative almost everywhere in $I.$

Thus, your assertion about the Weierstrass function fails --- your assertion implies differentiability almost everywhere and the Weierstrass function is differentiable nowhere.
Theorem 4.3.8 on p. 44 of [1] states the following (reworded slightly here), where $D^{+}f$ is the right upper Dini derivate of $f$:

Let $I$ be a nonempty open interval in $\mathbb R$ and let $f:I \rightarrow {\mathbb R}$ be continuous on $I$ and nowhere differentiable on $I$ (means: for each $x \in I,$ the 2-sided derivative of $f$ at $x$ does not exist finitely). Then $D^{+}f$ takes on every real value in every nonempty open subinterval of $I.$ In fact, more is true --- for each real number $r,$ the set $\{x \in I: \; D^{+}f(x) = r\}$ has cardinality $c.$ Note this easily implies the following stronger version --- for each real number $r$ and for each nonempty open subinterval $I_0$ of $I,$ the set $\{x \in I_{0}: \; D^{+}f(x) = r\}$ has cardinality $c$ (follows by applying the previous result to $f$ restricted to $I_{0}).$

Thus, your assertion about the Weierstrass function $W$ fails --- your assertion is that $D^{+}W(x) < 0$ for no $x,$ but we actually have $D^{+}W(x) < 0$ for continuum many $x.$ In fact, in every nonempty open interval we have $D^{+}W(x) = -2744$ for continuum many $x,$ and in every nonempty open interval we have $D^{+}W(x) = -10^{100}\sqrt{17}$ for continuum many $x,$ and in every nonempty open interval we have $\ldots$
[1] Andrew Michael Bruckner, Differentiation of Real Functions, 2nd edition, CRM Monograph Series #5, American Mathematical Society, 1994, xii + 195 pages. Review of 1980 1st edition (Note: Both of the results above are also in the 1980 1st edition, same chapter-section-item number, but the page numbers are different.)
A: This is what I can prove following the ideas of the original proof of Weierstrass. For a real number $r$ denote by $\langle r\rangle$ the unique integer such that
$$
-\frac12<r-\langle r\rangle\le\frac12.
$$
Let $x\in\mathbb{R}$.


*

*If there is a sequence of integers $m_k$ converging to $\infty$ such that $\langle b^{m_k}\,x\rangle$ is odd for all $k$, then
$$
\limsup_{h\to0+} \frac{f(x+h)-f(x)}{h}=+\infty.
$$

*If there is a sequence of integers $m_k$ converging to $\infty$ such that $\langle b^{m_k}\,x\rangle$ is even for all $k$, then
$$
\limsup_{h\to0-} \frac{f(x+h)-f(x)}{h}=+\infty.
$$
In this case case I have not been able to prove anything about the sign of the right upper Dini derivative.

