Proof: How many continuous/bounded functions on $[0,1]$ verify $f(x)=f(x/2)\frac{1}{\sqrt{2}}$?

Question:
How many continuous/bounded functions on $$[0,1]$$ verify $$f(x)=f(x/2)\frac{1}{\sqrt{2}}$$?

Answer:
Thank to @TonyK @Ryszard Szwarc. I think that i found an ever stronger demonstration that prooves that $$\exists ! f_0(x)=0 \; s.t. \; f(x)=f(x/2)\frac{1}{\sqrt{2}}$$ for all the functions bounded on $$[0;1]$$.
So i edited my first post.

I/ It's very easy to show that $$f_0(x)=0 \; \forall x \in [0;1]$$ verifies $$0=f(x)=f(x/2)\frac{1}{\sqrt{2}}=0$$

II/ Now by absurd we assume that it exists at least one bounded function $$f_1(x)$$ that is different to the zero function and that verifies:$$f(x)=f(x/2)\frac{1}{\sqrt{2}}$$ ?

1. By assumption, if such function exists that means: $$\exists 0 \leq x_1 \leq 1 \; s.t. f_1(x_1) \neq 0$$ (Without loose of generality $$f_1(x_1)>0$$)
2. Now let's build the two sequences $$u_n$$ and $$v_n$$ as follow: $$u_1=x_1; \; u_n=u_{n-1}/2=x_1/2^n$$ and $$v_1=f_1(x_1); \; v_n=f_1(u_n)=f_1(u_{n-1}/2)$$.
3. But let's recall that: $$f(x)=f(x/2)\frac{1}{\sqrt{2}} \Rightarrow \sqrt{2}f(x)=f(x/2)$$. In consequence: $$\exists 0 \leq x' \leq 1 \; s.t. \; v_1=(\sqrt{2})f_1(x') \; s.t. \; f_1(x') \neq 0$$ and so $$v_n={\sqrt{2}}f_1(u_n)={(\sqrt{2})}^nf_1(x') \;$$
4. It is obvious that $$u_n \underset{n\rightarrow \infty }{\rightarrow} 0$$ and that $$v_n \underset{n\rightarrow \infty }{\rightarrow} \infty$$. So we get that $$f_1(x)$$ is not bounded because it exists a sequence $$v_n$$ of elements of $$f_1(x)$$ that goes to infinity!

III/ And because we know from Weirstrass theorem that all continuous functions are bounded on a closed interval so

Q.E.D.

Is it correct?

• What is II supposed to be proving? Commented May 21, 2022 at 22:54
• @MarkSaving That if it exist $0 \leq x_0 \leq 1$ s.t. $f(x_0)=0$ so $f(x)=0$ for every x in [0;1]. (f(x)=the constante function zero). Btw i don't understant how the system of up and down vote of topic work on this site. Commented May 21, 2022 at 22:59
• I think you have it backwards: $f(x_n)=\sqrt 2 f(x_{n-1})$, not $\frac{1}{\sqrt 2} f(x_{n-1})$. So $f(x_n)$ tends to $\infty$, not $0$. Commented May 22, 2022 at 0:22
• @TonyK i don't understand what you re saying because: $y_1=f(x_1)=cste$ and so $y_n=y_1(2)^{(-n/2)}$ Commented May 22, 2022 at 9:20
• $y_0=f(x_0)=f(1)$. And putting $x=1$ gives $y_1=f(x_1)=f(x/2)=\sqrt 2f(x)=\sqrt 2y_0>y_0$. Continuing in this way, by induction we get $y_n=2^{n/2}y_0$. Is it clear now? Commented May 22, 2022 at 9:23

3 Answers

Let $$M=\displaystyle\sup_{0\le x\le 1}|f(x)|.$$ Then $$0\le M\le 2^{-1/2}\sup_{0\le x\le {1\over 2}}|f(x)|\le 2^{-1/2}M$$ Hence $$M=0,$$ i e. $$f(x)= 0$$ for any $$x\in [0,1].$$

Continuity is not essential. It suffices to assume that $$f$$ is bounded.

• Thank for your comment, i think that i found an other proof to this (thk to @TonyK) for bounded function. I posted it in the answer, and i will be happy to know if at your opinion it is correct? Commented May 22, 2022 at 11:05

Thank to @TonyK @Ryszard Szwarc and their comments. I think that i found an ever stronger demonstration that prooves that $$\exists ! f_0(x)=0 \; s.t. \; f(x)=f(x/2)\frac{1}{\sqrt{2}}$$ for all the functions bounded on $$[0;1]$$.

I/ It's very easy to show that $$f_0(x)=0 \; \forall x \in [0;1]$$ verifies $$0=f(x)=f(x/2)\frac{1}{\sqrt{2}}=0$$

II/ Now by absurd we assume that it exists at least one bounded function $$f_1(x)$$ that is different to the zero function and that verifies:$$f(x)=f(x/2)\frac{1}{\sqrt{2}}$$ ?

1. By assumption, if such function exists that means: $$\exists 0 \leq x_1 \leq 1 \; s.t. f_1(x_1) \neq 0$$ (Without loose of generality $$f_1(x_1)>0$$)
2. Now let's build the two sequences $$u_n$$ and $$v_n$$ as follow: $$u_1=x_1; \; u_n=u_{n-1}/2=x_1/2^n$$ and $$v_1=f_1(x_1); \; v_n=f_1(u_n)=f_1(u_{n-1}/2)$$.
3. But let's recall that: $$f(x)=f(x/2)\frac{1}{\sqrt{2}} \Rightarrow \sqrt{2}f(x)=f(x/2)$$. In consequence: $$\exists 0 \leq x' \leq 1 \; s.t. \; v_1=(\sqrt{2})f_1(x') \; s.t. \; f_1(x') \neq 0$$ and so $$v_n={\sqrt{2}}f_1(u_n)={(\sqrt{2})}^nf_1(x') \;$$
4. It is obvious that $$u_n \underset{n\rightarrow \infty }{\rightarrow} 0$$ and that $$v_n \underset{n\rightarrow \infty }{\rightarrow} \infty$$. So we get that $$f_1(x)$$ is not bounded because it exists a sequence $$v_n$$ of elements of $$f_1(x)$$ that goes to infinity!

III/ And because we know from Weirstrass theorem that all continuous functions are bounded on a closed interval so

Q.E.D.

Is it correct?

• In my opinion it is correct. Occasionally I use the proof by contradiction, but only when any other idea does not come to my mind. Commented May 22, 2022 at 12:04
• @RyszardSzwarc Thank you. Commented May 22, 2022 at 12:31

Your proof seems ok especially idea of part III. However it is hard to read due to lack of some kind of order. It suffices to see that function $$f$$ has to be bounded (certainly it is) since is continuous on compact set. Moreover it is straightforward that $$f(0)=0$$. Now we claim $$(\forall x\in[0,1])(\forall n\in\mathbb{N})f(x)=\frac{1}{\sqrt{2^n}}f\Big(\frac{x}{2^n}\Big).$$ The formal proof of the claim can be done by induction but I will leave as that. The final part can be done like that $$(\forall \epsilon>0)(\forall x\in[0,1]) |f(x)|<\epsilon.$$ Indeed let $$\epsilon>0$$ and $$x\in[0,1]$$ be fixed. Take $$n\in\mathbb{N}$$ large enough so the inequality $$\frac{\displaystyle \sup_{0\le x\le 1}|f(x)|}{\sqrt{2^n}}<\epsilon.$$ hold. So we have the following $$|f(x)|= \bigg|\frac{1}{\sqrt{2^n}}f\Big(\frac{x}{2^n}\Big)\bigg|\le\frac{\displaystyle \sup_{0\le x\le 1}|f(x)|}{\sqrt{2^n}}<\epsilon.$$ Since $$\epsilon$$ was arbitrary $$f(x)=0$$ for all $$x$$.