Difference of values of continuous function $f:\left[\frac{1}{2 \sqrt{2}},2\sqrt{2}\right]\to\mathbf{R}$ is continuous and $f\left(2\sqrt{2}\right)-f\left(\frac{1}{2 \sqrt{2}}\right)=3$. How do I show that for some $x$ in the domain $f(2x)-f(x)=1$?

Define $g(x) = f(2x) - f(x)$ for $\displaystyle x \in \left[\frac{1}{2\sqrt{2}}, \sqrt{2}\right]$. Note that $\displaystyle \frac{1}{2\sqrt{2}} = \frac{\sqrt{2}}{4}$, thus$$
g\left(\frac{\sqrt{2}}{4}\right) + g\left(\frac{\sqrt{2}}{2}\right) + g(\sqrt{2}) = f(2\sqrt{2}) - f\left(\frac{\sqrt{2}}{4}\right) = 3.
$$
If any of $\displaystyle g\left(\frac{\sqrt{2}}{4}\right)$, $\displaystyle g\left(\frac{\sqrt{2}}{2}\right)$ and $g(\sqrt{2})$ equals $1$ then it is already done. Otherwise, since their sum is $3$, they cannot all be less than $1$ or all be greater than $1$. Enumerate each possible scenario with respect to the relations of $\displaystyle g\left(\frac{\sqrt{2}}{4}\right)$, $\displaystyle g\left(\frac{\sqrt{2}}{2}\right)$, $g(\sqrt{2})$ and $1$, and apply the intermediate value theorem to prove the existence of the required $x$.
