11.2. Find all continuous functions $f: \mathbb{R} \rightarrow \mathbb{R}$, that satisfy the relation $f(x) = 5 f(5x) - 5x, \forall x \in \mathbb{R}$.
Solution. Consider the function $g: \mathrm{i} \rightarrow \mathrm{i}, g(x)=f(x)-\frac{5}{24} x, \forall x \in \mathrm{i}$. We have $g(5 x)=f(5 x)-\frac{25}{24} x, \forall x \in i$. Since $f(5 x)=\frac{1}{5} f(x)+x, \forall x \in \mathrm{i}$, it follows that $g(5 x)=\frac{1}{5} f(x)-\frac{1}{24} x=\frac{1}{5} g(x), \forall x \in i$, which implies the recurrence relation $g(x)=\frac{1}{5} g\left(\frac{1}{5} x\right), \forall x \in \mathbf{i}$, i.e., $g(x)=\frac{1}{5^{n}} g\left(\frac{1}{5^{n}} x\right), \forall x \in \mathrm{i}, \forall n \in \mathbf{N}$.

Since the function $f$ is continuous, it follows that the function $g$ is also continuous. We have

$$
g(x)=\lim _{n \rightarrow \infty} g(x)=\lim _{n \rightarrow \infty}\left(\frac{1}{5^{n}} g\left(\frac{1}{5^{n}} x\right)\right)=\lim _{n \rightarrow \infty}\left(\frac{1}{5^{n}}\right) \cdot g\left(\lim _{n \rightarrow \infty}\left(\frac{1}{5^{n}} x\right)\right)=0, \forall x \in i
$$

which implies $f(x)=\frac{5}{24} x, \forall x \in i$. Substituting into the initial relation, we obtain that this function is the only one that satisfies the condition of the problem.