Let $ R$ be the set of all real numbers. For $ g: R\to R\cup \{\pm\infty \}$ and $ x\in R$ , we define $ \limsup_{y\to x}g(y)\in R\cup \{\pm\infty \}$ in the usual way. Namely,

$ \limsup_{y\to x}g(y):=\inf_{\delta>0}\sup_{0<|y-x|<\delta}g(y)$ .

Next, for any $ f: R\to R$ , we define $ B_f\subset R$ as

$ B_f := \{ x\in R\mid \limsup_{y\to x} \bigg| \frac{f(y)-f(x)}{y-x} \bigg| <+\infty \}.$

Recently, I have realized that the following interesting claim (probably) holds:

Claim 1: Let $ f:R\to R$ and $ F_i\subset R~(i=1,2,3,\cdots)$ satisfy the following three conditions:

- Each $ F_i$ is a closed set.
- Each $ F_i$ has no interior points.
- $ R-B_f\subset \cup_i F_i$ .
(In other words, we assume that $ R-B_f$ is a set of first category.) Then there exists an open interval $ (a,b)\subset R$ and $ N\geq 1$ such that

$ \forall x,y\in (a,b)~~[~~|f(y)-f(x)|\leq N|y-x|~~]$ .

My proof is as follows. This proof is based on the Baire category theorem. Could anyone judge the correctness of this proof?

Proof of Claim 1: First, we can prove the following claim.

Claim 2: Let $ f:R\to R$ and $ x\in R$ satisfy $ \limsup_{y\to x} \bigg| \frac{f(y)-f(x)}{y-x} \bigg| <+\infty$ . Then there exist positive integers $ N,M\geq 1$ such that

$ \forall y,z\in R~~[~~x-\frac{1}{M}<y<x<z<x+\frac{1}{M}\Rightarrow |f(z)-f(y)|\leq N(z-y)~~]$ .

The proof of this claim 2 is basically the same as that of “straddle lemma” (so we omit the proof). Next, let $ f:R\to R$ and $ F_i\subset R~(i=1,2,3,\cdots)$ satisfy the three conditions of claim 1. For any positive integers $ N,M\geq 1$ , we define $ B_{N,M}\subset R$ as

$ B_{N,M}:=$ $ \{ x\in R\mid \forall y,z\in R~~[~~x-\frac{1}{M}<y<x<z<x+\frac{1}{M} \Rightarrow |f(z)-f(y)|\leq N(z-y)~~]~~\}$ .

By claim 2, we have $ B_f\subset \cup_{N,M\geq 1}B_{N,M}$ . Keeping in mind $ R-B_f\subset \cup_i F_i$ , we have

$ R\subset (\cup_{N,M\geq 1}B_{N,M})\cup (\cup_i F_i).~~~~$ (1)

Next, we show that each $ B_{N,M}$ is closed. Let $ x\in R$ and $ x_i\in B_{N,M}~(i\geq 1)$ satisfy $ x_i\to x~(i\to +\infty)$ . We have only to show that $ x\in B_{N,M}$ . Let $ y,z\in R$ satisfy

$ x-\frac{1}{M}<y<x<z<x+\frac{1}{M}$ .

Since $ x_i\to x$ , we have

$ x_i-\frac{1}{M}<y<x_i<z<x_i+\frac{1}{M}~$ (for sufficiently large $ i$ ).$ ~~~~$ (2)

Let $ i$ be one of them. Then it follows from (2) and $ x_i\in B_{N,M}$ that $ |f(z)-f(y)|\leq N(z-y)$ . Thus we conclude that

$ \forall y,z\in R~~[~~x-\frac{1}{M}<y<x<z<x+\frac{1}{M}\Rightarrow |f(z)-f(y)|\leq N(z-y)~~]$ .

This implies $ x\in B_{N,M}$ , so $ B_{N,M}$ is closed. Then the right hand side of (1) is a countable union of closed sets. Applying Baire category theorem, there exists $ F_i$ or $ B_{N,M}$ which has interior points. Since $ F_i$ has no interior points, it follows that there exists $ B_{N,M}$ which has interior points. So there exists an open interval $ (a,b)$ such that $ (a,b)\subset B_{N,M}$ . Without loss of generality, we may assume $ b-a<1/M$ and $ (a,b)\subset B_{N,M}$ . Next, let $ x,y\in (a,b)$ . We would like to show that $ |f(y)-f(x)|\leq N|y-x|$ . We may assume $ x\leq y$ . If $ x=y$ , then we have $ |f(y)-f(x)|\leq N|y-x|$ . If $ x<y$ , then let $ c:=(x+y)/2$ . Since $ b-a<1/M$ , we have

$ c-1/M<x<c<y<c+1/M.~~~~$ (3)

Moreover, we have $ c\in (x,y)\subset (a,b)\subset B_{N,M}$ , i.e. $ c\in B_{N,M}$ . Then it follows from (3) and $ c\in B_{N,M}$ that $ |f(y)-f(x)|\leq N(y-x)$ . Thus we complete the proof.$ ~~$ Q.E.D.

In addition, since the above proof proceeds in a conventional way, it is most likely that claim 1 is already known (if correct). So I am looking for the references which refer to the above claim 1. Does anyone know?