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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?

Whistleroosh
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    Your idea is fine. Just evaluate your expression at $\sqrt 2, \frac {\sqrt 2}2, \frac {\sqrt 2}4$ and sum. – lulu Jan 15 '22 at 19:19
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    If you find an $x$ such that $f(2x) - f(x) > 1$ and an $y$ such that $f(2y) - f(y) < 1$ then you can construct a $z$ such that $f(2z) - f(z) = 1$ by using the intermediate value theorem. – user3257842 Jan 15 '22 at 19:21
  • @lulu thanks but which expression should I evaluate? – Whistleroosh Jan 15 '22 at 19:43
  • Let $g(x)=f(2x)-f(x)$. Evaluate $g$ at the three points I mentioned. – lulu Jan 15 '22 at 19:44
  • but then I just get $g(\sqrt{2}) + g(\frac{\sqrt{2}}{2}) + g(\frac{\sqrt{2}}{4}) = 3$ – Whistleroosh Jan 15 '22 at 19:50
  • And what if you write that out in terms of $f$? Keep in mind, that you are assuming that either $g(x)>1$ or $g(x)<1$ for all relevant $x$. – lulu Jan 15 '22 at 19:58

2 Answers2

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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.

lulu
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  • Hmm, but what if $g(\sqrt{2}) > 1$ $g(\frac{\sqrt{2}}{2}) > 1$ $g(\frac{\sqrt{2}}{4}) < 1$? Then they can still add up to 3. Is it because of what @user3257842 said? It seems that what he said is true though I can't come up with a proof – Whistleroosh Jan 15 '22 at 20:21
  • I don't understand. If a continuous function on a closed interval is $<1$ at some point and $>1$ at another then it must be $1$ at some point. That's the usual Intermediate Value theorem. – lulu Jan 15 '22 at 20:23
  • Oh yes, thank you! I was thinking why what @user3257842 said is true for $f$ but it is actually easy to proof for $g$ – Whistleroosh Jan 15 '22 at 20:27
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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 : $$\exists x \in \left[a, 2^{n - 1} a\right], f(2 x) - f(x) = 1$$

Essaidi
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