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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.

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2 Answers 2

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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.

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  • $\begingroup$ I think they ended up with $7^n$ instead of $7^{-n}$. Is there some difference between what you did here compared to their solution? @daniel-geyfman $\endgroup$
    – Victor
    Aug 4, 2023 at 20:13
  • $\begingroup$ @Victor The $7^{n}$ in their solution results from the substition $x \to 7x+1$(at least I assume). In my solution, I use $x \to \frac{x}{7}-\frac{1}{7}$, which is the "inverse" of their substitution. So, $n$ substitutions of $x \to 7x+1$ is equivalent to "$-n$" substitutions of $x \to \frac{x}{7}-\frac{1}{7}$, which is where my $7^{-n}$ comes from. If you were to recreate my solution with their substitution, you would end up with a $7^{n}$ $\endgroup$
    – dgeyfman
    Aug 4, 2023 at 21:09
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Repeated application means apply the formula multiple times. For example if you apply it $2$ times you will have:

$$f''(x) = f''(7x+1) = f''(7(7x+1) + 1) = f\left(7^2 x + 8\right) = f\left(7^2 \left(x +\frac16\right) - \frac{7^2}6 + 8\right) = f\left(7^2 \left(x +\frac16\right) - \frac16\right)$$

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