Prove that if function $f$ is continuous on given interval and satisfies given equation there exists $x$ such that $f(2x) - f(x) = 1$ I am struggling to solve this question:

Function $f$ is continuous on interval $[\frac{1}{2\sqrt{2}}, 2\sqrt{2}]$ and $f(2\sqrt{2}) - f(\frac{1}{2\sqrt{2}}) = 3$. Show that there exists $x$ such that $f(2x) - f(x) = 1$

I didn't come up with any solution, but I was trying to do it in this way: 
Let's say $a = 2\sqrt{2}$ then $\frac{a}{8} = \frac{1}{2\sqrt{2}}$. Assume that there is no $x$ such that $f(2x) - f(x) = 1$. Then for every $x$ either $f(2x) - f(x) > 1$ or $f(2x) - f(x) < 1$. So $f(a) - f(\frac{a}{2}) \neq 1$ and $f(\frac{a}{2}) - f(\frac{a}{4}) \neq 1$ and $f(\frac{a}{4}) - f(\frac{a}{8}) \neq 1$. If the previous 3 equations were always $<$ or $>$ I could add them and I would get $f(a) - f(\frac{a}{2}) + f(\frac{a}{2}) - f(\frac{a}{4}) + f(\frac{a}{4}) - f(\frac{a}{8}) = f(a) - f(\frac{a}{8})$ which must be smaller than 3 or greater than 3 (depending on whether those 3 equations where always $<$ or $>$). Unfortunately some of those 3 equations may be $>$ and some $<$.
I don't know if I should keep trying with this idea or do something completely different. Could someone please help me?
 A: To follow the idea sketched in your post:  Define $g(x)=f(2x)-f(x)$.  Note that $g(x)$ is defined and continuous for $x\in \big [\frac {\sqrt 2}4,\sqrt 2\big ]$.
We want to show that $g(x)=1$ has a solution in the range.  Assume, to the contrary, that it is either strictly $>1$ or strictly $<1$ in the entire interval.
We note that $$g\left(\frac {\sqrt 2}4\right)=f\left(\frac {\sqrt 2}2\right)-f\left(\frac {\sqrt 2}4\right)$$
$$g\left(\frac {\sqrt 2}2\right)=f\left(\sqrt 2\right)-f\left(\frac {\sqrt 2}2\right)$$
$$g\left(\sqrt 2\right)=f\left(2\sqrt 2\right)-f\left(\sqrt 2\right)$$
Again, we are assuming that these all $>1$ or they are all $<1$.
Summing these we note that most terms cancel and we are left with $$f(2\sqrt 2)-f\left(\frac {\sqrt 2}4\right)$$
And we conclude that this must be $>3$ or $<3$, contradicting the assumption that it is, in fact, $3$. And we are done.
A: We can generalize this exercise :
let $a \in \mathbb{R}$, $n \in \mathbb{N}^*$ and $f : \left[a, 2^n a\right] \to \mathbb{R}$ a continuous function such that :
$$f \left(2^n a\right) - f \left(a\right) = n$$
Show that :
$$\exists x \in \left[a, 2^{n - 1} a\right], f(2 x) - f(x) = 1$$
