Functional Equation: A Simple One.

With $y=-x$ we see that $f(0)^2=f(0)f(2x)$ for all $x$ , so either $f$ is constant or $f(0)=0$ . If $f(0)=0$ then, putting $y=x$ we see that $f(x)^2=0$ for all $x$ . In this case $f$ is constant and zero. Thus, without needing continuity, the set $\mathcal {F}$ consists of the constant functions, so has the same cardinality as $\mathbb{R}$ .

If we let $y=0$ , we get $\left(f\left(\frac x2\right)\right)^2=f(x)f(x)$ , which implies $\left|f\left(\frac x2\right)\right|=\left|f(x)\right|$ We will demonstrate by contradiction that $f$ only satisfies this and is continuous if $f$ is constant.

Assume there are two positive real numbers $a$ and $b$ such that $|f(a)|\neq|f(b)|$ . By the above equation, $|f(a)|=\left|f\left(\frac a2\right)\right|=\left|f\left(\frac a4\right)\right|=\left|f\left(\frac a8\right)\right|=\dots$ $|f(b)|=\left|f\left(\frac b2\right)\right|=\left|f\left(\frac b4\right)\right|=\left|f\left(\frac b8\right)\right|=\dots$ Now consider $\displaystyle\lim_{x\to0^+}f(x)$ . Since the sequences $a,\frac a2, \frac a4,\dots$ and $b,\frac b2, \frac b4,\dots$ approach $0$ , given any positive $\delta$ there will always be two numbers $\frac a{2^n}$ and $\frac b{2^m}$ such that they are less than $\delta$ . Since $|f(\frac a{2^n})|=|f(a)|$ and $|f(\frac b{2^m})|=|f(b)|$ , $\left|f\left(\frac a{2^n}\right)-f\left(\frac b{2^m}\right)\right|\geq\big| |f(a)|-|f(b)|\big|$ by the reverse triangle inequality. $\big| |f(a)|-|f(b)|\big|$ is nonzero and constant, hence as $x$ approaches $0$ , there are values of $f(x)$ at least this distance apart, and the limit does not exist. Hence, the function is not continuous.

So, $f$ can only be continuous if for all positive reals $a$ and $b$ , $|f(a)|=|f(b)|$ . By the mean value theorem we can then conclude that $f(a)=-f(b)$ is also prohibited unless $f$ is always zero, or else we would require $f(x)=0$ at some point $x$ between $a$ and $b$ , in which case $|f(x)|\neq |f(a)|$ . Hence, $f(a)=f(b)$ for all positive reals $a$ and $b$ . We can then repeat the above process for negative reals, taking the limit as x approaches zero from the negative side. Then, the limit as x approaches zero only exists if it is equal from both sides, so $f(a)=f(b)$ for all real numbers $a$ and $b$ .

Thus, $f$ is constant. It is easy to check that all constant functions satisfy the equation in the question, and so the cardinality of $\mathcal F$ is equal to the cardinality of $\mathbb R$ .

Taking $y=x$ , we have $f(x)^2=f(2x)\;f(0)$

If $f(0)=0$ , this tells us $f(x)=0$ for all $x$ . Now assume $f(0)=a \neq 0$ .

Taking $y=-x$ , we have $\begin{aligned} f(0)^2 &=f(2x)\;f(0) \\ a &=f(2x) \end{aligned}$

so that $f(x)=a$ for all $x$ . It's clear that this function satisfies the original equation for any real $a$ ; hence the cardinality of $\mathcal{F}$ is equal to the cardinality of $\mathbb{R}$ .

Let us assume that $f(0)=0.$ Then for any real number $t$ making $x=t$ and $y=t$ into the functional equation, we get $\left(f(t)\right)^2=f(2t) f(0)=0.$ So in this case we get a possible solution that is $f=0.$ It is easy to check that it is a solution of the given functional equation. In a similar way, we can also prove that if $f$ is a solution of the functional equation and there is a number $a$ such that $f(a)=0,$ then $f=0.$ Therefore, any solution $f$ that is different from 0, must satisfy the condition that $f(x)\neq 0$ for any real number $x.$ A continuous function satisfying this condition, have to have a constant sign over the whole set of real numbers.

Now, let us assume that $f(0)=c,$ where $c$ is any real number different from zero. Replacing $y$ by 0 into the functional equation, we obtain that for any real number $x$ the following equation holds $\left (f(\frac{x}{2})\right )^2=(f(x))^2.$ Then $|f(x)|=|f(\frac{x}{2})|,$ and as $f$ has constant sign, then $f(x)=f(\frac{x}{2}).$ By repeated use of the last equation, we can prove that $f(x)=f(\frac{x}{2^n})$ for every real number $x$ and every natural number $n.$ Taking limits in the previous equation as $n \rightarrow \infty$ and using the continuity of the function, we obtain that $f(x)=c.$ Therefore, $f$ is a constant function.

Then any continuous solution of the given functional equation has to be constant and it is easy to prove that any constant function defined on the set of all real numbers has to be a solution. So $\mathcal{F}$ would be formed by all constant functions defined on the set of all real numbers. Therefore, the cardinality of $\mathcal{F}$ is equal to the cardinality of $\mathbb{R}.$