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:$f(x)=f(x/2)\frac{1}{\sqrt{2}}$ ?

*

*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 $f_1(x_1)>0$)

*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 $v_1=f_1(x_1); \; v_n=f_1(u_n)=f_1(u_{n-1}/2)$.

*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') \; $

*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?
 A: 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.
A: 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:$f(x)=f(x/2)\frac{1}{\sqrt{2}}$ ?

*

*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 $f_1(x_1)>0$)

*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 $v_1=f_1(x_1); \; v_n=f_1(u_n)=f_1(u_{n-1}/2)$.

*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') \; $

*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?
A: 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 $x$.
