gpt pro 5.4 attempt on Math-657, version 7

Problem 50: derivatives of the totient distribution function (consolidated progress report)

Problem statement and final status

Let \[ G(c):=\lim_{X\to\infty}\frac1X\#\left\{n\le X:\frac{\varphi(n)}n\le c\right\} \qquad(0\le c\le 1). \] By continuity of $G$ (see the references below), this is the same function as the density \[ f(c)=\lim_{X\to\infty}\frac1X\#\{n\le X:\varphi(n)<cn\}. \] The original Erdős problem asks whether there exists no point $x\in[0,1]$ such that \[ G'(x)=f'(x) \] exists and is strictly positive.

This note consolidates all rigorous advances established in the thread. It does not claim a full proof or disproof of the original problem. The present status is therefore \[ \boxed{\text{UNRESOLVED}}. \]

Verified literature input used throughout

We use the following standard facts.

1. (Schoenberg) The limiting distribution function $G$ exists, is continuous, and is strictly increasing on $[0,1]$.

2. (Erdős) The measure \[ \mu:=dG \] is purely singular with respect to Lebesgue measure $\lambda$. In particular, \[ G'(x)=0\qquad\text{for }\lambda\text{-a.e. }x\in(0,1). \]

3. (Tenenbaum--Toulmonde; Toulmonde) Write \[ \mathcal E:=\left\{\frac{\varphi(n)}n:n\ge1\right\}. \] For \[ K(t):=\sum_{\varphi(m)/m=t}\frac1m, \qquad \mathcal A(\varepsilon):=\left\{\frac{\varphi(n)}n:1\le n\le \varepsilon^{-1/8}\right\}, \] and \[ L(x):=\exp\!\left(\sqrt{\log x\,\log\log x}\right), \] one has, uniformly for $0<\varepsilon<1/3$ and $t\in\mathcal A(\varepsilon)$, \[ G(t)-G(t-\varepsilon t)=K(t)\left(G(1)-G(1-\varepsilon)\right)+O\left(L(1/\varepsilon)^{-1/5}\right), \] \[ G(t+\varepsilon t)-G(t)\ll L(1/\varepsilon)^{-1/5}, \] and, as $\varepsilon\to0^+$, \[ G(1)-G(1-\varepsilon)=\frac{e^{-\gamma}}{\log(1/\varepsilon)}+O\!\left(\frac1{(\log(1/\varepsilon))^2}\right). \]

Everything below is either deduced from these facts or proved directly in the thread.

Basic structure of the value set $\mathcal E$

1. Density and uniqueness of prime support

Proposition 1. The set \[ \mathcal E=\left\{\frac{\varphi(n)}n:n\ge1\right\} =\left\{\prod_{p\mid n}\left(1-\frac1p\right):n\ge1\right\} \] is countable and dense in $[0,1]$. Moreover, if $P,Q$ are finite sets of primes and \[ \prod_{p\in P}\left(1-\frac1p\right)=\prod_{q\in Q}\left(1-\frac1q\right), \] then $P=Q$.

Proof. Countability is immediate. For density, let $u\in(0,1)$ and write $y:=-\log u>0$. Since $\sum_p 1/p$ diverges and $1/p\to0$, a standard greedy subseries argument provides distinct large primes $p_1,\dots,p_k$ with \[ \sum_{i=1}^k\frac1{p_i}=y+o(1). \] Because \[ \log\left(1-\frac1p\right)=-\frac1p+O\left(\frac1{p^2}\right) \] and $\sum_p 1/p^2<\infty$, we obtain \[ \sum_{i=1}^k\log\left(1-\frac1{p_i}\right)=-y+o(1), \] so \[ \prod_{i=1}^k\left(1-\frac1{p_i}\right)=u\,e^{o(1)}\to u. \] Thus $\mathcal E$ is dense in $(0,1)$, and clearly also accumulates at $0$ and $1$.

For uniqueness, rewrite the equality as \[ \prod_{p\in P}\frac{p}{p-1}=\prod_{q\in Q}\frac{q}{q-1}. \] Let $r=\max(P\cup Q)$. The $r$-adic valuation of the left-hand side is $1$ if $r\in P$ and $0$ otherwise, because $r$ cannot divide any denominator factor $p-1$ with $p<r$. The same holds on the right with $Q$. Hence $r\in P$ iff $r\in Q$. Cancelling the common factor $r/(r-1)$ and descending on the largest prime gives $P=Q$. $\square$

2. Closed form for $K(t)$

Proposition 2. Let $t\in\mathcal E$. Write uniquely \[ t=\prod_{p\in P}\left(1-\frac1p\right) \] with $P$ a finite prime set. Then \[ K(t)=\sum_{\varphi(m)/m=t}\frac1m=\prod_{p\in P}\frac1{p-1}. \] In particular, $K(t)>0$ for every $t\in\mathcal E$.

Proof. The value $\varphi(m)/m$ depends only on the prime support of $m$, and Proposition 1 shows that the prime support is uniquely determined by $t$. Therefore $\varphi(m)/m=t$ exactly when the set of prime divisors of $m$ is $P$. Hence \[ K(t)=\prod_{p\in P}\sum_{\nu\ge1}\frac1{p^\nu} =\prod_{p\in P}\frac{1/p}{1-1/p} =\prod_{p\in P}\frac1{p-1}. \] $\square$

Sharp left-hand asymptotics at all special values $t\in\mathcal E$

Theorem 3. For every $t\in\mathcal E$, \[ G(t)-G(t-h)=\frac{e^{-\gamma}K(t)+o(1)}{\log(t/h)} \qquad(h\to0^+). \] Equivalently, \[ \frac{G(t)-G(t-h)}{h} \sim \frac{e^{-\gamma}K(t)}{h\log(t/h)} \qquad(h\to0^+), \] so the left derivative exists in the extended sense and equals $+\infty$: \[ G'_-(t)=+\infty \qquad(t\in\mathcal E). \]

Proof. Fix $t\in\mathcal E$ and choose $n_0\ge1$ with $t=\varphi(n_0)/n_0$. For every sufficiently small $\varepsilon>0$ we have $n_0\le\varepsilon^{-1/8}$, hence $t\in\mathcal A(\varepsilon)$. The literature formula therefore gives \[ G(t)-G(t-\varepsilon t)=K(t)\left(G(1)-G(1-\varepsilon)\right)+O\left(L(1/\varepsilon)^{-1/5}\right). \] Since \[ G(1)-G(1-\varepsilon)=\frac{e^{-\gamma}}{\log(1/\varepsilon)}+O\!\left(\frac1{(\log(1/\varepsilon))^2}\right) \] and \[ \log(1/\varepsilon)\,L(1/\varepsilon)^{-1/5}\to0, \] we obtain \[ G(t)-G(t-\varepsilon t)=\frac{e^{-\gamma}K(t)+o(1)}{\log(1/\varepsilon)}. \] Now put $h=\varepsilon t$, so that $\log(1/\varepsilon)=\log(t/h)$. This yields the first asymptotic. Dividing by $h$ gives the second, and since $h\log(t/h)\to0^+$, the quotient tends to $+\infty$. $\square$

Corollary 4. If there exists $x\in(0,1)$ such that $G'(x)$ exists and $0<G'(x)<\infty$, then necessarily \[ x\notin\mathcal E. \]

Measure-theoretic consequences of singularity

1. Symmetric upper densities

Proposition 5. Let $\nu$ be a finite Borel measure on $\mathbb R$ with $\nu\perp\lambda$. Then \[ \limsup_{h\to0^+}\frac{\nu((x-h,x+h))}{2h}=+\infty \qquad\text{for }\nu\text{-a.e. }x. \] Applied to $\mu=dG$, this gives \[ \limsup_{h\to0^+}\frac{G(x+h)-G(x-h)}{2h}=+\infty \qquad\text{for }\mu\text{-a.e. }x\in(0,1). \]

Proof. For $M>0$, let \[ A_M:=\left\{x:\limsup_{h\to0^+}\frac{\nu((x-h,x+h))}{2h}\le M\right\}. \] Choose a Borel set $S$ with $\lambda(S)=0$ and $\nu(\mathbb R\setminus S)=0$. Fix $\eta>0$ and choose an open set $O\supset S$ with $\lambda(O)<\eta$. For each $x\in A_M\cap S$, there exists $r_x>0$ such that \[ \nu((x-r,x+r))\le 2(M+1)r\qquad(0<r<r_x), \] and, after shrinking $r_x$, also $(x-r_x,x+r_x)\subset O$. A one-dimensional Besicovitch covering argument produces a countable cover by such intervals with overlap bounded by $2$. Summing the measure bounds over that subfamily yields \[ \nu(A_M\cap S)\le 4(M+1)\lambda(O)\le 4(M+1)\eta. \] Letting $\eta\to0$ gives $\nu(A_M)=0$. Since this holds for every $M$, the claim follows. $\square$

2. One-sided upper densities

Proposition 6. Let $\nu$ be a finite Borel measure on $\mathbb R$ with $\nu\perp\lambda$. Then \[ \limsup_{h\to0^+}\frac{\nu((x-h,x])}{h}=+\infty, \qquad \limsup_{h\to0^+}\frac{\nu([x,x+h))}{h}=+\infty \] for $\nu$-almost every $x$. Applied to $\mu=dG$, this gives \[ \overline D_-G(x)=\overline D_+G(x)=+\infty \qquad\text{for }\mu\text{-a.e. }x\in(0,1), \] where \[ \overline D_-G(x):=\limsup_{h\to0^+}\frac{G(x)-G(x-h)}{h}, \qquad \overline D_+G(x):=\limsup_{h\to0^+}\frac{G(x+h)-G(x)}{h}. \] In particular, \[ \mu\left(\{x:\overline D_-G(x)=+\infty\}\right)= \mu\left(\{x:\overline D_+G(x)=+\infty\}\right)=1. \]

Proof. The proof is the same as in Proposition 5, using one-sided intervals $(x-r,x]$ or $[x,x+r)$ instead of symmetric intervals. $\square$

Category-theoretic consequences: residual oscillation

For $x\in(0,1)$ define the one-sided lower Dini derivatives \[ \underline D_-G(x):=\liminf_{h\to0^+}\frac{G(x)-G(x-h)}{h}, \qquad \underline D_+G(x):=\liminf_{h\to0^+}\frac{G(x+h)-G(x)}{h}. \] Also define the symmetric quotient \[ S_h(x):=\frac{G(x+h)-G(x-h)}{2h} \qquad(0<h<\min\{x,1-x\}). \]

1. Residual symmetric flatness and blow-up

Proposition 7. The sets \[ A_{\mathrm{sym}}:=\left\{x\in(0,1):\liminf_{h\to0^+}S_h(x)=0\right\}, \qquad B_{\mathrm{sym}}:=\left\{x\in(0,1):\limsup_{h\to0^+}S_h(x)=+\infty\right\}, \] are both dense $G_\delta$ subsets of $(0,1)$. Moreover, $A_{\mathrm{sym}}$ has full Lebesgue measure, and $B_{\mathrm{sym}}$ has full $\mu$-measure.

Proof. For each fixed rational $q>0$, the map $x\mapsto S_q(x)$ is continuous on its natural domain, so both sets admit standard rational-scale $G_\delta$ descriptions. Since $G'(x)=0$ for $\lambda$-a.e. $x$, the symmetric quotient tends to $0$ for $\lambda$-a.e. $x$, whence $A_{\mathrm{sym}}$ has full measure and therefore is dense. Proposition 5 implies that $B_{\mathrm{sym}}$ has full $\mu$-measure; since every open interval has positive $\mu$-measure (because $G$ is strictly increasing), $B_{\mathrm{sym}}$ is also dense. $\square$

2. Residual one-sided oscillation

Proposition 8. The sets \[ A_-:=\{x\in(0,1):\underline D_-G(x)=0\}, \qquad A_+:=\{x\in(0,1):\underline D_+G(x)=0\} \] are dense $G_\delta$ subsets of $(0,1)$ and each has full Lebesgue measure. The sets \[ B_-:=\{x\in(0,1):\overline D_-G(x)=+\infty\}, \qquad B_+:=\{x\in(0,1):\overline D_+G(x)=+\infty\} \] are dense $G_\delta$ subsets of $(0,1)$; in fact $\mu(B_-)=\mu(B_+)=1$.

Proof. Each set is $G_\delta$ by the same rational-scale continuity argument as above.

If $x\in A_{\mathrm{sym}}$, then there exists $h_n\to0^+$ with \[ \frac{G(x+h_n)-G(x-h_n)}{2h_n}\to0. \] But \[ \frac{G(x+h_n)-G(x-h_n)}{2h_n} = \frac12\frac{G(x)-G(x-h_n)}{h_n} + \frac12\frac{G(x+h_n)-G(x)}{h_n}, \] and both summands are nonnegative because $G$ is increasing. Hence each summand tends to $0$, so $x\in A_-\cap A_+$. Thus $A_{\mathrm{sym}}\subseteq A_-\cap A_+$, and therefore $A_-$ and $A_+$ are dense and of full Lebesgue measure.

For $B_-$, every $t\in\mathcal E$ satisfies $G'_-(t)=+\infty$ by Theorem 3, so $\mathcal E\subseteq B_-$. Since $\mathcal E$ is dense, $B_-$ is dense.

For $B_+$, fix a nonempty open interval $I=(a,b)$ and integers $N,m\ge1$. Choose $t\in\mathcal E\cap(a+(b-a)/3,b-(b-a)/3)$ and then $h\in(0,1/m)$ so small that \[ \frac{G(t)-G(t-h)}{h}>N. \] By continuity of the left secant quotient as a function of the scale, the same inequality holds at some rational $q\in(0,1/m)$ near $h$. Setting $x:=t-q$ gives $x\in I$ and \[ \frac{G(x+q)-G(x)}{q}=\frac{G(t)-G(t-q)}{q}>N. \] Thus every basic open set in the rational-scale $G_\delta$ description of $B_+$ is dense, so $B_+$ is dense.

Finally, the $\mu$-full statements follow from Proposition 6. $\square$

3. A residual set of maximal one-sided oscillation outside $\mathcal E$

Theorem 9. Define \[ \Omega:=A_-\cap A_+\cap B_-\cap B_+\cap\left((0,1)\setminus\mathcal E\right). \] Then $\Omega$ is a dense $G_\delta$ subset of $(0,1)$, and for every $x\in\Omega$ one has \[ \underline D_-G(x)=\underline D_+G(x)=0, \qquad \overline D_-G(x)=\overline D_+G(x)=+\infty. \] In particular, $G$ is not differentiable at any point of $\Omega$.

Proof. By Proposition 8, the four sets $A_-,A_+,B_-,B_+$ are dense $G_\delta$ subsets of $(0,1)$. Since $\mathcal E$ is countable, its complement is also dense $G_\delta$. Their intersection is therefore dense $G_\delta$, and the stated Dini-derivative identities hold by construction. Any finite derivative would force all four one-sided Dini derivatives to agree finitely, contradiction. $\square$

Exact local secant spectra on a residual set

For $x\in(0,1)$ and $0<h<\min\{x,1-x\}$, define \[ Q_h^-(x):=\frac{G(x)-G(x-h)}{h}, \qquad Q_h^+(x):=\frac{G(x+h)-G(x)}{h}. \] For fixed $x$, both functions of $h$ are continuous on their natural domains.

Theorem 10. Let $x\in\Omega$. Then for every $c>0$ and every \[ 0<\delta<\min\{x,1-x\} \] there exist $h_-,h_+\in(0,\delta)$ such that \[ Q_{h_-}^-(x)=c, \qquad Q_{h_+}^+(x)=c. \] Equivalently, every positive slope is realized exactly by arbitrarily small left secants and by arbitrarily small right secants based at $x$.

If \[ \Sigma_-(x):=\left\{L\in[0,\infty):\exists h_n\to0^+,\ Q_{h_n}^-(x)\to L\right\}, \] \[ \Sigma_+(x):=\left\{L\in[0,\infty):\exists h_n\to0^+,\ Q_{h_n}^+(x)\to L\right\}, \] then \[ \Sigma_-(x)=\Sigma_+(x)=[0,\infty) \qquad(x\in\Omega). \]

Proof. Fix $x\in\Omega$, $c>0$, and $\delta$ as above. Since $\underline D_-G(x)=0$, there exists $a\in(0,\delta)$ with $Q_a^-(x)<c$. Since $\overline D_-G(x)=+\infty$, there exists $b\in(0,\delta)$ with $Q_b^-(x)>c$. By continuity of $h\mapsto Q_h^-(x)$, the intermediate value theorem yields $h_-\in(0,\delta)$ with $Q_{h_-}^-(x)=c$. The right-hand statement is identical.

Applying this for each $c>0$ and a sequence $\delta_m\to0^+$ yields $c\in\Sigma_-(x)\cap\Sigma_+(x)$. Since $\underline D_-G(x)=\underline D_+G(x)=0$, the value $0$ is also attained as a limit point. Nonnegativity of the quotients shows that no negative values occur. Hence the spectra are exactly $[0,\infty)$. $\square$

Interval consequences

Corollary 11. Every nonempty open interval $I\subset(0,1)$ contains subintervals with arbitrarily small average slope and subintervals with arbitrarily large average slope: \[ \inf_{\substack{a<b\\ [a,b]\subset I}}\frac{G(b)-G(a)}{b-a}=0, \qquad \sup_{\substack{a<b\\ [a,b]\subset I}}\frac{G(b)-G(a)}{b-a}=+\infty. \]

Proof. The lower statement follows by choosing $x\in A_{\mathrm{sym}}\cap I$ and taking $h>0$ small with $[x-h,x+h]\subset I$ and $S_h(x)$ arbitrarily small. The upper statement follows by choosing $t\in\mathcal E\cap I$ and using Theorem 3. $\square$

Corollary 12. Every nonempty open interval $I\subset(0,1)$ contains secants of every prescribed positive slope. More precisely, for every $c>0$ there exist points \[ a<x<b, \qquad x\in I\setminus\mathcal E, \qquad [a,b]\subset I, \] such that \[ \frac{G(x)-G(a)}{x-a}=c, \qquad \frac{G(b)-G(x)}{b-x}=c, \qquad \frac{G(b)-G(a)}{b-a}=c. \]

Proof. Choose $x\in\Omega\cap I$ and let $\delta>0$ be so small that $[x-\delta,x+\delta]\subset I$. Apply Theorem 10 to find $h_-,h_+\in(0,\delta)$ with $Q_{h_-}^-(x)=Q_{h_+}^+(x)=c$. Then $a:=x-h_-$ and $b:=x+h_+$ have the required properties, and the slope identity over $[a,b]$ follows by adding the two equal increments. $\square$

Final reduction of the candidate set

Define \[ D_+:=\{x\in(0,1):G'(x)\text{ exists and }0<G'(x)<\infty\}. \] Also set \[ \Xi:=\left(B_-\cap B_+\right)\setminus\mathcal E. \]

Proposition 13. The set $\Xi$ is dense $G_\delta$ in $(0,1)$ and satisfies \[ \mu(\Xi)=1. \] Moreover, \[ D_+\subseteq (0,1)\setminus\Xi. \] Hence every $x\in D_+$ satisfies all of the following:

1. $x\notin\mathcal E$;
2. both one-sided upper Dini derivatives at $x$ are finite;
3. $x$ lies in a meager set;
4. $x$ lies in a $\mu$-null set;
5. $x$ lies in a Lebesgue-null set.

In addition, \[ D_+\subseteq (0,1)\setminus\Omega, \] so any candidate point must also avoid the residual exact-spectrum set of Theorem 10.

Proof. The density and $G_\delta$ property of $\Xi$ follow from Proposition 8 and countability of $\mathcal E$. The equality $\mu(\Xi)=1$ follows from Proposition 6 together with $\mu(\mathcal E)=0$. If $G'(x)$ exists finitely, then both one-sided upper Dini derivatives are finite, so $x\notin B_-\cap B_+$, hence $x\notin\Xi$. The remaining assertions follow immediately. $\square$

What has been proved, and what remains open

The thread established the following rigorous picture.

1. The special value set $\mathcal E=\{\varphi(n)/n:n\ge1\}$ is countable and dense in $[0,1]$.
2. At every $t\in\mathcal E$ there is a sharp left asymptotic \[ G(t)-G(t-h)\sim \frac{e^{-\gamma}K(t)}{\log(t/h)}, \] so $G'_-(t)=+\infty$.
3. The upper symmetric Dini derivative is infinite for $\mu$-almost every point.
4. Both one-sided upper Dini derivatives are infinite for $\mu$-almost every point.
5. There is a dense $G_\delta$ set $\Omega\subset(0,1)\setminus\mathcal E$ such that for every $x\in\Omega$, \[ \underline D_-G(x)=\underline D_+G(x)=0, \qquad \overline D_-G(x)=\overline D_+G(x)=+\infty, \] and in fact every positive one-sided secant slope occurs exactly, on both sides, at arbitrarily small scales around $x$.
6. Every nonempty open interval contains subintervals with arbitrarily small average slope, subintervals with arbitrarily large average slope, and indeed secants of every prescribed positive slope.
7. Any hypothetical point $x$ with $0<G'(x)<\infty$ must lie outside $\mathcal E$, outside a residual exact-spectrum set, outside a residual and $\mu$-full one-sided blow-up set, and inside a set that is simultaneously meager, $\mu$-null, and Lebesgue-null.

These are strong obstructions, but they do not settle the original Erdős problem. No gap-free proof or counterexample has been obtained in this thread.

References

1. I. J. Schoenberg, On asymptotic distributions of arithmetical functions, Trans. Amer. Math. Soc. 39 $1936$, 315--330.

2. P. Erdős, On the distribution function of additive functions, Ann. of Math. $2$ 47 $1946$, 1--20.

3. G. Tenenbaum and V. Toulmonde, Sur le comportement local de la répartition de l'indicatrice d'Euler, Funct. Approx. Comment. Math. 35 $2006$, 321--338.

4. V. Toulmonde, Module de continuité de la fonction de répartition de $\varphi(n)/n$, Acta Arith. 121 $2006$, no. 4, 367--402.

5. P. Mattila, Geometry of Sets and Measures in Euclidean Spaces, Cambridge Studies in Advanced Mathematics 44, Cambridge Univ. Press, 1995.

6. T. F. Bloom, Erdős Problem #50, current discussion/status page, accessed 2026-03-09.