This is a continuation of my answer to Smallest positive zero of Weierstrass nowhere differentiable function
If $x=a$ is a knot point of a continuous function $f(x),$ then $\lim_{h \rightarrow 0}\frac{f(a+h) \, - \, f(a)}{h}$ fails to exist in the following way: The set of "sub-sequential limits" of $\frac{f(a+h) \, - \, f(a)}{h}$ as $h \rightarrow 0$ on the left is the entire extended real line $[-\infty, \, +\infty],$ and the set of "sub-sequential limits" of $\frac{f(a+h) \, - \, f(a)}{h}$ as $h \rightarrow 0$ on the right is the entire extended real line $[-\infty, \, +\infty].$ (See Theorem V in Vaidyanathaswamy's paper below.) This "sub-sequential limit" result can fail if $f$ is not continuous. Consider $a=0$ for the function defined by $g(x) = 1$ if $x$ is irrational, $g(x) = -1$ if $x$ is nonzero and rational, and $g(0) = 0.$ In this case, each of the unilateral sets of "sub-sequential limits" of $g$ at $0$ equals the two-point set $\{-\infty, \, +\infty\}.$
A cusp of $f(x)$ is a point $x=a$ such that both left and right derivatives of $f$ at $x=a$ exist infinitely with opposite signs. That is, $D_{-}f(a) = D^{-}f(a) = -\infty$ and $D_{+}f(a) = D^{+}f(a) = +\infty,$ or $D_{-}f(a) = D^{-}f(a) = +\infty$ and $D_{+}f(a) = D^{+}f(a) = -\infty.$
From p. 115: Such points are called cusps, and if this subset of cups exists [this means the set of cusps is not empty], its measure is known to be zero. Non-differentiable functions are known to exist, however, which have no cusps. Peano's function [= space filling curve's coordinate functions] and Singh's generalisation of it are examples. [Note: In fact, the set of cusps (of any function, not just of a continuous function) is countable. Even more, for a continuous function they form a rather small countable set in the sense that they form a scattered set in the Cantor-Bendixson sense. This is because they form a countable $G_{\delta}$ set. All these results follow from stronger results that William H. Young had already proved by 1909.]
Theorem 1 (p. 115): If $\xi$ is a zero of any non-differentiable function $f(x),$ and if $\xi$ is a knot-point, [then] $\xi$ is necessarily a limiting point of the zeros of $f(x).$
Corollary (p. 116): More generally, if $\xi$ is any knot-point of $f(x),$ [then] the equation $f(x) = f(\xi)$ has $\xi$ as a limiting point of its roots. In other words, there are an infinite number of roots of the equation $f(x) = f(\xi)$ in any interval enclosing $\xi.$
Theorem 2 (p. 116): The zeros of Weierstrass's non-differentiable function, other than those that may happen to be cusps, are limiting points of the zeros.
From p. 116: A similar result [to Theorem 2] at once follows for the equation $\phi(x) = \phi(\xi)$ where $\xi$ is any point other than a cusp of $\phi(x).$ [$\phi(x) = \sum a^n\cos(b^n\pi x)$ is the Weierstrass function.] Ganesh Prasad's result that $x = \frac{1}{2}$ is a limiting point of the zeros follows at once, since $x = \frac{1}{2}$ is not a cusp. It is possible that a cusp can be a zero of $\phi(x) \, \ldots$
From p. 117: (1 footnote omitted) Lastly, it must be noted that a zero which may occur at a cusp is necessarily isolated. For, suppose [$\cdots$] Theorem 2 will evidently be true for any non-differentiable function for which the property proved by G. C. Young for Weierstrass's function can be extended. [Note: The specific property Srinivasiengar is referring to is the property that at each non-cusp point there exists a side (left or right) such that on that side both $-\infty$ and $+\infty$ are Dini derivates and on the other side at least one of the Dini derivates is infinite.] A consideration of Peano's function will however show that this property is not true for all non-differentiable functions.
[9] Ramaswamy Vaidyanathaswamy (1894-1960), On continuous functions of a real variable, Proceedings of the Indian Academy of Sciences. Section A 9 #1 (January 1939), 67-71. Zbl 21.01502; JFM 65.0195.03
This paper appears to have been written to provide a more modern mathematical approach to some of the questions raised in previous Indian work on nowhere differentiable continuous functions. Vaidyanathaswamy was the author of the first topology text in India, Treatise on Set Topology. Part I (Part II never appeared), published by the Indian Mathematical Society (in Madris) in August 1947. [The 2nd edition, Set Topology, was published by Chelsea Publishing Company in 1960, and the 2nd edition was reprinted by Dover Publications in 1999.] Most of Vaidyanathaswamy's research (about 30 of his 52 research papers) was in algebraic geometry, all published before 1940. During this time he also published papers in number theory, combinatorics, and determinants. Beginning in the late 1930s his interests shifted to partially ordered sets, Boolean algebras, lattices and their applications to general topology and logic, and he published 7 or 8 papers on these topics. Several Indian mathematicians who later worked on these topics were introduced to them in classes that Vaidyanathaswamy taught during his later years.
For biographies of Vaidyanathaswamy, see: (1) pp. i-iii (by Viakalathur Sankrithi Krishnan) in Collected Papers of Prof. R. Vaidyanathaswamy, University of Madras Press, 1957, vii + 589 pages; (2) Amur Narasinga Rao and V. Ganapathy Iyer, R. Vaidyanathaswamy (1894-1960), The Mathematics Student 29 (1961), 1-14 [reprinted in Journal of the Indian Mathematical Society (N.S.) 64 (1997), 1-13]; (3) pp. 12-16 & 46 (by Rao/Iyer) in J. N. Kapur, Some Eminent Indian Mathematicians of the Twentieth Century, 2nd edition, Mathematical Sciences Trust Society, 1991, 49 pages (1st edition published in 1983).
First two paragraphs of the paper: (1 footnote omitted) In a recent paper "On the Zeroes [sic] of Weierstrass'[s] Non-differentiable Function", Dr. C. N. Srinivasiengar proves that if a knot-point of Weierstrass' function be a zero of the function, it must necessarily be a limit point of zeroes. The principle involved in this result is elucidated in Theorems VI and VII of this paper, which obtain a statement of it in a more general form. Dr. Srinivasiengar also raises certain questions regarding the set of zeroes of Weierstrass' function, and quotes Dr. A. N. Singh's statement that 'it is desirable to find some special characteristic of the set of zeroes of Weierstrass' function, as for instance, whether it is closed or open, enumerable or not'. The answer to one at least of these questions is easy and well known. Namely, the set of points at which a continuous function $f(x)$ takes an assigned value $\alpha$---or briefly the set $S(\alpha)$ of the $\alpha$-points of $f(x)$---is necessarily a closed set. I concern myself here with these questions, in so far as they relate to the general continuous function.
Notation issues: Unless otherwise stated, $f$ is a continuous real-valued function defined on the closed interval $[0,1].$ The minimum and maximum values of $f$ [= lower and upper bounds of $f$] are denoted by $L$ and $U,$ and $\alpha$ denotes a real number such that $L \leq \alpha \leq U.$ The set of solutions to the equation $f(x) = \alpha$ is denoted by $S(\alpha).$
Theorem I (p. 67): $S(\alpha)$ is a closed set. Theorem II (p. 68): [If $f$ is not differentiable at each point in a set that is dense in $[0,1],$ then] $S(\alpha)$ is closed and non-dense [= nowhere dense]. Theorem III (p. 68): [For any continuous function] The set $S(\alpha)$ is of measure zero, except possibly for an enumerable set of values of $\alpha$ in [the closed interval] $(L,U).$ [Note: In this paper the same notation is used for open and closed intervals.]
Vaidyanathaswamy's proof of Theorem III is a standard technique, one that is often used when proving that a monotone function has at most countably many discontinuities or when proving that the sum (defined as the least upper bound of the finite subsums) of an uncountable set of positive real numbers is infinite. Specifically, he observes that for each $\epsilon > 0$ there exist at most finitely many values of $\alpha$ such that the measure of $S(\alpha)$ is greater than $\epsilon.$ Incidentally, since the proof only makes use of the fact that the sets $S(\alpha)$ are measurable, the same result holds for an arbitrary measurable function. [Measurability is needed so that we can make use of the fact that if $\alpha \neq \alpha',$ which implies (even without measurability) that we have $S(\alpha) \cap S(\alpha') = \emptyset,$ then the measure of $S(\alpha) \cup S(\alpha')$ is the sum of the measure of $S(\alpha)$ and the measure of $S(\alpha').]$ Vaidyanathaswamy does not mention the measurable generalization, but I am fairly certain that he was aware of it (both from what I know about him and from the structure of his proof).
On p. 69 Vaidyanathaswamy claims that the zero set of a nowhere differentiable continuous function $f:[0,1] \rightarrow {\mathbb R}$ can have any specified measure less than $1$ (including measure $0$). The claim is correct, but Vaidyanathaswamy's construction is not correct. Vaidyanathaswamy first correctly outlines the construction of a symmetric Cantor set in $[0,1]$ having an arbitrarily specified measure less than $1,$ and then he places an appropriate horizontally-scaled (but not vertically-scaled) copy of a Weierstrass type function in each of the bounded complementary intervals so that the scaled copy is zero at the endpoints of the interval and is positive at each point in the interior of the interval, and finally he defines the function to be zero at all points in the Cantor set. In carrying this out, Vaidyanathaswamy defines $f$ by putting $f(x) = \sum_{m=0}^{\infty} a^m \sin \left( b^{m} \pi \cdot \frac{x - {\alpha}_n}{{\beta}_n - {\alpha}_n} \right),$ where $ab > 1 + \frac{3\pi}{2}$ and ${\alpha}_n < x < {\beta}_{n},$ which defines $f$ for values of $x$ within the $n$'th complementary interval $({\alpha}_n, \, {\beta}_n).$ This construction does give a nowhere differentiable function whose zero set is the specified Cantor set of positive measure. However, the function is not continuous. In fact, the function is discontinuous at each point in the Cantor set. This is because each point in the Cantor set (note these are points where the value of the function is zero) is a limit of a sequence of complementary intervals, and the function has the same positive maximum value in each of the complementary intervals. This may have been an oversight or an editing issue. However, even if we consider the function $g$ defined by $g(x) = ({\beta}_n - {\alpha}_n)f(x)$ for $x$ in the complementary interval $({\alpha}_n, \, {\beta}_n),$ and $g(x) = 0$ for $x$ in the Cantor set (which may have been what Vaidyanathaswamy intended), the function $g$ will not work. The function $g$ will be continuous, but unfortunately $g$ will also have a zero derivative at almost all (Lebesgue measure) points in the Cantor set. Indeed, it is not difficult to see that $g$ has a zero derivative at every point where the Lebesgue density of the Cantor set is $1.$ (In fact, $g$ has a zero derivative on the possibly larger set of points at which the Cantor set does not have positive upper porosity.) For a correct construction of a continuous nowhere differentiable function on an arbitrarily specified closed nowhere dense set, see Lipinski's 1966 paper On zeros of a continuous nowhere differentiable function.
From p. 69: I shall next proceed to study the relation of the sets $S(\alpha)$ to the concept of 'derivate,' which is really the subject-matter of Dr. Srinivasiengar's result. Theorem IV (p. 69): If $\phi(h)$ is any function whatever [not necessarily continuous], defined in the open interval $(0,k),$ [then] the set of limits of $\phi(h)$ as $h \rightarrow +0$ is a closed set. [Note: This set of limits is (now) often called the right cluster set of $\phi$ at $0.$] Theorem V (p. 70): If the $\phi(h)$ of Theorem IV is continuous in the open interval $(0,k),$ [then] the closed set of limits of $\phi(h)$ as $h \rightarrow +0$ is a closed interval [in the extended real line]. [Note: The original had $h \rightarrow 0,$ but from the proof it is clear that this was a typo and $h \rightarrow +0$ was intended. Also, Vaidyanathaswamy includes the following footnote at the end of the statement of Theorem V: For theorems of this sort in Analysis, it is convenient to consider the real number continuum as a closed interval with the endpoints $+\infty$ and $-\infty.$] Theorem VI (p. 70) If in Theorem V, $t$ is an interior point of the closed interval of functional limits of $\phi(h),$ [then] we can find two sequences $(h_n),$ [and] $(h'_{n})$ [both] tending to $+0,$ such that $[\phi(h_n)]$ tends to $(t+0),$ and $\phi(h'_{n})$ tends to $(t-0).$ [Note: This means that if $t$ is an interior point of the right cluster set of $\phi,$ then $t$ is approached both from above and from below by values of $\phi$.] If $t$ is an endpoint of the interval of functional limits, [then] it may or may not be possible to find two sequences $(h_n)$, [and] $(h'_{n})$ with this property.
From pp. 70-71: We now apply theorems V and VI to the function $\phi(h)$ which is equal to the incrementary ratio $\frac{f(x+h) \, - \, f(x)}{h}$ of the continuous function $f(x)$ at the point $x.$ This function $\phi(h)$ is continuous in open intervals on either side of the endpoint $h=0.$ By Theorem V, the functional limits of $\phi(h)$ as $h \rightarrow +0$ and $-0,$ constitute respectively two closed intervals. I shall call these intervals the right and left derivate-intervals of the continuous function $f(x)$ at the point $x.$ Suppose now that $f(x) = \alpha,$ so that $x$ belongs to the closed set $S(\alpha).$ If a right [open] neighbourhood $(x, \, x+h)$ of $x$ can be found, throughout which $f(x) > \alpha$ $(< \alpha),$ [then] we say that $f(x)$ is increasing (decreasing) on the right of $x.$ If in every small right neighbourhood $(x, \, x+h)$ [we have] $f(x) \geq \alpha$ $(\leq \alpha),$ [then] we say that $f(x)$ is non-decreasing (non-increasing) on the right of $x.$ If in every small right neighbourhood $(x, \, x+h)$ of $x$ there are points at which $f(x) > \alpha$ as well as points at which $f(x) < \alpha,$ [then] we say that $f(x)$ is oscillatory on the right of $x.$ It is obvious that if $0$ be external to the right derivate interval of $f(x)$ at $x,$ then $f(x)$ must be either increasing or decreasing to the right of $x.$ If however $0$ is an endpoint of the right derivate-interval, all possibilities are open for the behaviour of $f(x)$ on the right of $x.$ If $x$ is a limitpoint of the set $S(\alpha),$ approached say from the right, [then] it is clear that $0$ must belong to the right derivate-interval. Is the converse of this theorem true? Theorem VI shows that the converse can be asserted categorically, only if it is known that $0$ is an interior point of the derivate-interval.
Theorem VII (p. 71): If $0$ is an interior point of the right derivate-interval of $f(x)$ at $x,$ then $f(x)$ is necessarily oscillatory on the right of $x.$ Also $x$ is necessarily a limitpoint of $S(\alpha)$ approached from the right, where $\alpha = f(x).$ Vaidyanathaswamy's proof: Since, by hypothesis, $0$ is an interior point of the derivate-interval, it follows from Theorem VI, that the incrementary ratio can approach zero from either side as $h \rightarrow +0.$ Hence $f(x)$ must be oscillatory to the right of $x.$ Hence in every right neighbourhood of $x,$ there are points at which $f(x) > \alpha,$ as well as points at which $f(x) < \alpha.$ Hence, since $f$ is continuous, there are in every right neighbourhood of $x,$ points at which $f(x) = \alpha;$ in other words $x$ is a limitpoint of $S(\alpha)$ approached from the right.
The remaining of the paper that follows the proof of Theorem VII: The point $x$ is said to be a knot-point of $f(x),$ if both the derivate-intervals at $x$ extend from $-\infty$ to $+\infty.$ In such a case zero is an interior point of both the derivate-intervals. Hence a knot-point $x$ of $f$ is necessarily a limitpoint on both sides of the set $S(\alpha)$ $[\alpha = f(x)]$---which is Dr. Srinivasiengar's result. It is clear that a result corresponding to Theorem VII can be stated for any other value of the derivate.
[10] Subbaramiah [Subramiah] Minakshisundaram (1913-1968), On the roots of a continuous non-differentiable function, Journal of the Indian Mathematical Society (N.S.) 4 #1 (March 1940), 31-33. MR 1,303a; Zbl 23.02001; JFM 66.1224.02
Note: In this paper "non-differentiable" means at each point neither a finite nor an infinite two-sided derivative exists, and $(a,b)$ denotes the closed interval $[a,b].$
First few paragraphs of the paper (pp. 31-32): (3 footnotes omitted) Let $f(x)$ be a continuous non-differentiable function defined in the interval $(0,1),$ whose upper and lower bounds are $l$ and $u,$ say. If $l \leq \alpha \leq u,$ we denote by $S(\alpha),$ the set of points $x$ in $(0,1)$ for which $f(x) = \alpha.$ It is known that $S(\alpha)$ is closed and non-dense [= nowhere dense; Vaidyanathaswamy's paper is cited in a footnote]. We now divide the interval $(l,u)$ into three distinct [= disjoint] sets of points $A,$ $B$ and $C,$ so that $A + B + C = (l,u),$ [$+$ denotes union] and (a) $A=$ the set of points $\alpha$ for which $S(\alpha)$ is of positive measure (b) $B=$ the set of points $\alpha$ for which $S(\alpha)$ is non-enumerable but of measure zero (c) $C=$ the set of points $\alpha$ for which $S(\alpha)$ is enumerable. We have then the following main theorem of this note: THEOREM: For any continuous non-differentiable function $f(x),$ (1) $A$ is at most enumerable, (2) $B$ has the measure of the interval $(l,u),$ (3) $C$ is of measure zero at most. In other words a continuous non-differentiable function takes almost every value more than enumerably infinitely often. It is known that in the case of Singh's function $A$ and $C$ are both empty. In the case of Weierstrass's non-differentiable function $W(x) = \sum_{n=0}^{\infty}a^n \cos b^nx\pi$ where $0<a<1,$ [and] $b$ is an odd integer and $ab > 1 + \frac{3\pi}{2}(1-a),$ I show that $S(l)$ and $S(u)$ are enumerable, so that $C$ is not empty. Also it is shown that $C$ can include only the proper maxima and minima, so that $C$ is at most enumerable. We now proceed to the proof of the main theorem. That $A$ is at most enumerable has been already observed by Dr. R. Vaidyanathaswamy. [Note: This is Theorem III in Vaidyanathaswamy's 1939 paper.]
Minakshisundaram next observes that (2) will now follow after (3) is proved, and he proves (3) by making use of the following result from p. 277 of Theory of the Integral (1937) by Stanislaw Saks. Lemma (p. 32): Suppose that $F$ is continuous, and that $E$ is a set at no point of which the function $F$ has a derivative (finite or infinite). Suppose further that each point $x$ of $E$ is an isolated point of the set $E_{t}[F(t) = F(x)].$ Then $|E| = |F(E)| = 0.$ [Note: $E_{t}[F(t) = F(x)]$ means $\{t \in E: \; F(t) = F(x)\}.]$ Proof of Lemma: Here $|E|$ denotes the measure of $E$ and $F(E)$ is the set of the values of $F$ over $E.$ Now to every point $\alpha$ of $C,$ we associate a point $x = x_{\alpha}$ so that (i) $f(x_{\alpha})=\alpha$ (ii) $x_{\alpha}$ is an isolated point of the set $S(\alpha).$ The choice (ii) is always possible, as $S(\alpha),$ being closed and enumerable, must contain [at least one] isolated points. We shall denote by $X$ the set of points $x_{\alpha}$ so chosen as to satisfy (i) and (ii). Now by the lemma, as $f(x)$ is non-differentiable, $|f(X)| = |X| = 0.$ But $C = f(X)$ Hence $|C|=0.$
After the proof of the Lemma, Minakshisundaram remarks that for the Weierstrass function the set $C$ is countable. Finally, the following remark from the last paragraph of the paper seems incorrect to me: A similar remark applies to any continuous non-differentiable function $f(x).$ In general $S(\alpha) = P_{\alpha} + I_{\alpha},$ where $P_{\alpha}$ is a perfect set and $I_{\alpha}$ is composed of isolated points. Minakshisundaram's remark seems to be incorrect because any closed nowhere dense set can be the zero set of a nowhere differentiable (finitely or infinitely) continuous function [Lipinski, On zeros of a continuous nowhere differentiable function, 1966; replace Lipinski's use of the van der Waerden function with the Weierstrass function], and there exist such sets that do not have the form of a perfect set union an isolated set, such as the set $\{0\} \cup \{\frac{1}{2}, \, \frac{1}{3}, \, \frac{1}{4}, \, \ldots \}.$ It is possible that Minakshisundaram intended "In general" to convey some type of "typically but not necessarily always" meaning, but even when read in context in the last paragraph, it does not seem to me that a "typically but not necessarily always" meaning was intended.
[11] Zygmunt Zahorski (1914-1998), Sur l'ensemble des racines de l'équation $W(x) = f(x)$ [On the set of roots of the equation $W(x) = f(x)],$ Comptes Rendus des Séances de la Société des Sciences et des Lettres de Varsovie [Varsovie = Warsaw]. Classe III 41 (1948), 43-45. MR 13,219a; Zbl 39.05604
In this paper $W(x)$ is the Weierstrass function. In what follows I have rewritten some of the results from their original wording (when translated to English), but I have preserved the original terminology and notation. Theorem I (p. 43): Let $\alpha(x)$ and $f(x)$ be continuous functions such that, at each point, both (a) and (b) are true: (a) $-\infty$ and $+\infty$ are among the values of the Dini derivates of $\alpha(x)$ at that point; (b) all $4$ Dini derivates of $f(x)$ are finite at that point. In addition, assume that $[\alpha(u) - f(u)][\alpha(v) - f(v)] < 0.$ [That is, at one endpoint of the interval $[u,v]$ the graph of $y = \alpha(x)$ is above the graph of $y = f(x),$ and at the other endpoint of this interval the graph of $y = \alpha(x)$ is below the graph of $y = f(x)\,).]$ Then $\alpha(x) = f(x)$ for continuum many $x$ in the open interval $(u,v).$ Zahorski observes that $\alpha(x)$ can be the Weierstrass function $W(x),$ and then he observes that Theorem I follows from Theorem II.
Theorem II (p. 43): Let $b(x)$ be a continuous function such that, at each point, at least one of the Dini derivates of $b(x)$ at that point is positive and at least one of the Dini derivates of $b(x)$ at that point is negative. In addition, assume that $b(c)b(d) < 0.$ [That is, the function $b(x)$ has opposite-signed values at the endpoints of the interval $[c,d].]$ Then $b(x) = 0$ for continuum many $x$ in the open interval $(c,d).$ Zahorski proves Theorem II by introducing the idea of a crossing point of a function. We say that $x_0$ is a crossing point of $b(x)$ if, for each neighborhood of $x_0,$ there exist points $x_{-}$ and $x_{+}$ in this neighborhood such that $b(x_{-}) < 0$ and $b(x_{+}) > 0.$ Let $E$ be the set of crossing points of $b(x)$ that belong to the open interval $(c,d).$ Zahorski observes that $E$ is closed and he proves that $E$ is dense in itself, and hence $E$ is a perfect set. Zahorski also observes that $E$ contains all the zeros of $b(x)$ in the open interval $(c,d).$ Therefore, the set of zeros of $b(x)$ in the open interval $(c,d)$ contains a nonempty perfect set, and hence this set of zeros has cardinality continuum.
Corollary (p. 44): For each real number $a$ such that $\min W(x) < a < \max W(x),$ the equation $W(x) = a$ has continuum many solutions. Zahorski ends the paper by observing that Minakshisundaram showed $W(x) = a$ has continuum many solutions for all but countably many values of $a$ in the open interval $(\min W(x), \, \max W(x)).$ Thus, Zahorski's corollary strengthens Minakshisundaram's "all but countably many" for the Weierstrass function to "all but two". Incidentally, because the level sets are closed, being uncountable is equivalent to having cardinality continuum. However, in real analysis literature during the first half of the 20th century authors frequently did not state the more precise "cardinality continuum" property, even those authors in which it is clear they knew it held.
Translation of last sentence of the paper: The set $\{cx+b = {\alpha}(x)\}$ has cardinality of the continuum for each function $cx+b,$ where $c \neq 0$ and ${\alpha}(x) \equiv W(x),$ [hence] the construction of the function ${\alpha}(x)$ (Mr. Erdös and Mr. Gillis) is superfluous. Zahorski does not give a reference in this case, but the reference is almost certainly to a function constructed in Gillis' 1939 paper Note on a conjecture of Erdös. The first sentence of the Gillis paper is: In this paper I construct a continuous periodic function of $x$ such that the curve $y = f(x)$ intersects any non-vertical straight line, if at all, in an infinite number of points. This is not possible. In fact, for each continuous function $f$ there exist co-countably many values of $c \in \mathbb R$ such that at least one line with slope $c$ intersects the graph of $y = f(x)$ in exactly one point. [See Bruckner's 1994 book Differentiation of Real Functions (Chapter 13.4, p. 146, Theorem 4.2(i)) or see Bruckner/Garg's 1977 paper The level structure of a residual set of continuous functions (p. 315, Theorem 4.3(a)).] This does not contradict Zahorski's Theorem I (because for each such line, the points $u$ and $v$ do not exist), but it does contradict what seems to be Zahorski's intended application of Theorem I.
[12] K. Padmavally (??-??), On the roots of equation $f(x) = \xi$ where $f(x)$ is real and continuous in $(a,b)$ but monotonic in no subinterval of $(a,b)$, Proceedings of the American Mathematical Society 4 #6 (December 1953), 839-841. MR 15,513h; Zbl 52.05201
Let $f:[a,b] \rightarrow {\mathbb R}$ be continuous and let $m,\,M$ be the minimum, maximum values of $f$ on the interval $[a,b].$ For each $a \leq \xi \leq b,$ let $S(\xi) = \{x: \; f(x) = \xi\}.$ Theorem 1 states that if $f$ is monotone in no interval, then $S(\xi)$ has cardinality continuum for a second category set of values of $\xi.$ Since older authors sometimes said "second category" when "complement of first category" was proved (mostly in the 1910s to the 1930s), I mention that Padmavally only proves "second category". However, it is true that $f(\xi)$ has cardinality continuum for residually many $\xi \in [m,M]$ (i.e. all but a first category set of $\xi \in [m,M]).$ This stronger version was proved by Solomon Marcus in 1958, and an even stronger version was proved by K. M. Garg in 1963, who proved "nonempty perfect" in place of "cardinality continuum". For a still further strengthening, see Theorem 7 in Brown/Darji/Larsen's 1999 paper Nowhere monotone functions and functions of nonmonotonic type.
Padmavally mentions that Minakshisundaram had proved that if $f$ is nondifferentiable, then $f(\xi)$ has cardinality continuum for almost all (Lebesgue measure) $\xi \in [m,M].$ [Incidentally, Padmavally's description of this result is slightly ambiguous, because in Minakshisundaram's paper "nondifferentiable" means "nowhere a finite or infinite derivative", and slightly ahistorical, because in Minakshisundaram's paper "uncountable" was used instead of "cardinality continuum".] Padmavally notes that being monotone in no interval is much weaker than being nowhere (finitely) differentiable, and in fact, there exist functions monotone in no interval that are finitely differentiable at each point.
Character posting limitations (30,000 characters, including spaces, per answer) prevent me from saying anything about the following two papers. However, in my opinion these two papers best follow those that I have already discussed, both from a mathematical perspective and from a historical perspective.
[13] Solomon Marcus (1925- ), Sur les fonctions continues qui ne sont monotones en aucun intervalle [On continuous functions that are not monotone in any interval], Académie de la République Populaire Roumaine. Revue de Mathématiques Pures et Appliquées [after 1963: Revue Roumaine de Mathématiques Pures et Appliquées] 3 (1958), 101-105. MR 21 #6403; Zbl 87.27402
[14] Krishna Murari Garg (1932- ), On level sets of a continuous nowhere monotone function, Fundamenta Mathematicae 52 (1963), 59-68. MR 26 #1405; Zbl 106.26904
Best Answer
Assume WLOG that $\phi(x)>0$ when $x>0$. Since the limit described exists for all $x$ in the source of $f$. We get for any $x$ the bound:
$f(x+\delta)-f(x) \leq C\phi(\delta)$
for $0 < \delta < \delta_0$ for some $C,\delta_0>0$ which may depend on $x$.
diving by $\delta$ we get by the assumptions on $\phi$ that
$\underline{\lim}_{\delta \to 0} ( \frac{f(x+\delta) - f(x)}{\delta}) \leq 0$
This is one the four derivatives of $f$, and proposition 2 chapter 5 in Real Analysis by H.L. Royden states that if $f$ is continuous then it is (non-strictly) decreasing. Similar for increasing. So $f$ is constant.