Anti-symmetric derivative implies differentiability

Question

Wikipedia defines Symmetric derivative as a limit $$\lim_{h\to 0} \frac{f(x+h)-f(x-h)} {2h}$$ which doesn't imply differentiability. An obvious example is $f(x)=|x|$ or $f=\chi_{\mathbb Q}$

I think a slightly different, unbalanced form of the limit guarantees differentiability. Especially the following limit $$\lim_{h\to 0} \frac{f(a+2h)-f(a-h)} {3h}=\gamma$$ and the continuity of $f$ at $a$ implies $f'(a)=\gamma$.

My question is,

1. Is the following proof valid?
2. Is it possible to prove an analogous extension with the condition $$\lim_{h\to 0} \frac{f(x+ph)-f(x-qh)} {(p+q)h} = \gamma$$ while $p,q$ are different positive real values?

(Proof)

Define $$G(h)=\begin{cases} {\frac{f(a+2h)-f(a-h)} {3h}-\gamma}\quad h\neq 0 \\ 0 \quad\quad\quad \quad \quad\quad\quad\quad h=0\\ \end{cases} $$

and substitute $h$ with $\frac 12 (-\frac 12)^kh = x_k h$.

$$G(x_kh)=\frac{f(a+ (-\frac 12)^kh)-f(a+(-\frac 12)^{k+1}h)} {3x_k h}-\gamma$$

By multiplying $3x_k$ both side and adding up from $k=0$ to $n-1$,

$$3\sum_{k=0}^{n-1}{x_k G(x_k h)}=\sum_{k=0}^{n-1}(\frac{f(a+ (-\frac 12)^kh)-f(a+(-\frac 12)^{k+1}h)} h)- (1-(-\frac 12)^n)\gamma $$

RHS become $$\frac{f(a+h)-f(a-(1/2)^n)} h - (1-(-\frac 12)^n)\gamma\quad \to \quad\frac{f(a+h)-f(a)} h -\gamma$$ by continuity.

So we have to show that the left side is zero. Since $|G(x)|<M$ for some $0$ neighborhood, Weierstrass M-test guarantees that $\sum_k ^\infty{x_k G(x_kh)}$ is continuous at $h=0$ and obviously its value is zero.

Answer 1

I think that your proof is correct, and also works for the generalized assumption. Since $(p, q)$ can be scaled by a common factor, it can be stated as follows:

Let $I \subset \Bbb R$ be an interval and $f: I \to \Bbb R$ a function which is continuous at $a \in I$. If $$\lim_{h \to 0} \frac{f(a+h)-f(a+ch)}{(1-c)h} = \gamma$$ for some $c \in (-1, 1)$ then $f$ is differentiable at $a$, and $f'(a) = \gamma$.

For the proof I would proceed as follows: To simplify the notation one can assume that $a=0$, $f(0) = 0$, and $\gamma = 0$ so that $$\tag{*}\lim_{h \to 0} \frac{f(h)-f(ch)}{(1-c)h} = 0 \, .$$

Then for all $n > 0$ $$ \frac{f(x)}{x} = (1-c)\sum_{k=0}^{n-1} c^k \frac{f(c^kx)-f(c^{k+1}x)}{c^k x - c^{k+1}x} + \frac{f(c^nx)}{x} $$ From the condition $(*)$ we can conclude that for a given $\varepsilon > 0$, $$ \left|\frac{f(x)}{x} \right| \le (1-c)\sum_{k=0}^{n-1} c^k \varepsilon + \left|\frac{f(c^nx)}{x} \right| \le \varepsilon + \left|\frac{f(c^nx)}{x} \right| $$ for sufficiently small $x \ne 0$. For fixed $x$ and $n \to \infty$ it follows that $$ \left|\frac{f(x)}{x} \right| \le \varepsilon $$ because of the continuity of $f$ at $x=0$. Since $\varepsilon > 0$ was arbitrary, we have proved that $$ f'(0) = \lim_{x \to 0}\frac{f(x)}{x} = 0 \, . $$