Find all functions $f : \mathbb{R} \to \mathbb{R}$ that have continuous second derivative and for which the equality $f(7x+1)=49f(x)$ holds for all $x \in \mathbb{R}$.
I found a solution online to the problem and I want to ask about a part of the solution.
The solution is the following:
Solution. Differentiating twice gives that $f''(7x + 1) = f''(x)$. Repeatedly applying this equation (in both directions), we get that $f''(x) = f''(7^n (x + 1/6) - 1/6)$ for all $n \in \mathbb{Z}$. Since $f''$ is continuous, letting $n \rightarrow -\infty$, we get $f''(x) = f''(-1/6)$, and thus $f''(x)$ is constant for all $x \in \mathbb{R}$. This immediately implies that $f$ is quadratic, and we can then directly substitute in to find that the set of solutions is $$ f(x) = ax^2 + \frac{1}{3}a x + \frac{1}{36}a, $$for any $a \in \mathbb{R}$.
what I wanted to ask about is what is meant by this "repeated application of the equation"? I'm not very familiar with recursion, but I suspect that there is some recursive argument going on with this.
It is a bit weird that they leave this part out, it seems pretty non-obvious.
Let's look at the equation $$f''(7x+1)=f''(x)$$
Say we substitute $\frac{x}{7}-\frac{1}{7}$ for $x$. This will give us $$f''(7(\frac{x}{7}-\frac{1}{7})+1)=f''(\frac{x}{7}-\frac{1}{7})$$ $$f''(x)=f''(\frac{x}{7}-\frac{1}{7})$$
Let's do the same substitution again. We get $$f''(\frac{x}{7}-\frac{1}{7})=f''(\frac{x}{49}-\frac{8}{49})$$ But we know that $f''(x)=f''(\frac{x}{7}-\frac{1}{7})$ so we have $$f''(x)=f''(\frac{x}{49}-\frac{8}{49})$$
If we perform the same substitution over and over again to infinity(with the same reduction back to $f''(x)=...$), we will end up with $$f''(x)=f''(\frac{-1}{7} + \frac{1}{7}(\frac{-1}{7}+\frac{1}{7}(\frac{-1}{7}+\cdots)))$$ To evaluate this, we can exploit self-similarity. Let $y=\frac{-1}{7} + \frac{1}{7}(\frac{-1}{7}+\frac{1}{7}(\frac{-1}{7}+\cdots))$. Then, we can notice that $y$ appears again in the expression and substitute as follows: $$y=\frac{-1}{7}+\frac{1}{7} y$$ $$\frac{6}{7}y=\frac{-1}{7}$$ $$y=\frac{-1}{6}$$ and hence $f''(x)=f''(\frac{-1}{6})$ and the solution proceeds as in OP.
If you want to rigorously define "repeating substitution infinitely many times," then you can show that after $n$ substitutions the equation you get is $$f''(x)=f''(7^{-n}(x+1/6)-1/6)$$ and then you take the limit as $n \to \infty$. You will end up with $$f''(x)=f''(\frac{-1}{6})$$ and again the solution proceeds as in OP.
Note: here, I only substitute in one direction, but by doing the reverse substitution $x \to 7x+1$ instead of $x \to \frac{x}{7}-\frac{1}{7}$ you get a similar equation.