Find $f:\mathbb{R}\to\mathbb{R}$, such that $f$ is continuous at $0$ and $f(2x)-f(x)=x^2$.

Question

Find $f:\mathbb{R}\to\mathbb{R}$, such that $f$ is continuous at $0$ and $f(2x)-f(x)=x^2$ for all $x\in\mathbb{R}$.

Answer 1

Given $$f(2x)-f(x)=x^2$$ we can form $$f(x)-f(x/2)=\left(\frac{x}{2^1}\right)^2\\ f(x/2)-f(x/4)=\left(\frac{x}{2^2}\right)^2\\ .\\.\\.\\ f(x/2^{n-1})-f(x/2^n)=\left(\frac{x}{2^n}\right)^2$$

Adding above equations telescopically as $n\to\infty$ gives $$\begin{align} f(x)&=f(0)+x^2\cdot\frac{(\frac{1}{2})^2}{1-(\frac{1}{2})^2}\\ &=\frac{1}{3}\cdot x^2 +f(0) \end{align}$$

Answer 2

Hint: Write $f(x)=f(x/2) + x^2/4$ and iterate: $f(x)=x^2(1/4+1/16) + f(x/4) =...=x^2/3 + \lim_{k\rightarrow \infty} f(x/2^k) = x^2/3+c$ by continuity at 0 ($c=f(0)$ any real number).

Answer 3

Suppose $f(x) = ax^2+bx+c$. Your condition requires $$a(2x)^2+b(2x)+c-ax^2-bx-c=x^2$$ $$(3a-1)x^2+bx=0$$ for all $x$. This happens if and only if both $3a-1=0$ and $b=0$, i.e., if $a=\frac13$ and $b=0$ (note $c$ is arbitrary). So any function of the form $$f(x) = \frac13x^2 + c$$ will do (of course, there may be others).

Addendum: If you suppose $f(x) = \sum_{n=0}^{\infty}a_nx^n$, then the condition requires $$\sum_{n=0}^{\infty}a_n(2x)^n-\sum_{n=0}^{\infty}a_nx^n = x^2$$ $$\sum_{n=0}^{\infty}(2^n-1)a_nx^n = x^2$$ $$0a_0 + a_1x + (3a_2-1)x^2 + \sum_{n=3}^{\infty}(2^n-1)a_nx^n = 0$$ This implies that $a_0$ may be chosen arbitrarily, that $a_1=0$, that $a_2=\frac13$, and that $(2^n-1)a_n=0$ for $n\geq 3$ (so $a_n=0$ for $n\geq 3$ since $2^n-1\neq 0$ for $n\geq 3$). So the only functions of this form are, as before, of the form $f(x) = \frac13x^2 + a_0$.

Answer 4

The solution i found is $$ f(x)=\frac{1}{3}x^2+c$$ i don't know if it's unique or not, tell me if you need more details.

Explanation:

I proceeded as follows:

First, i assumed that $f\in C^1(\mathbb{R})$ and that $f'$ is a linear function.

The first assumption is because i wanted to calculate the derivative of the given equation.

Then, we obtain: $$2f'(2x)-f'(x)=2x $$ Using the linearity of $f'$ we write, $$2f'(2x)-\color{red}{\frac{1}{2}}f'(\color{red}2x)=2x $$ Thus, $$\frac{3}{2}f'(2x)=2x $$ $$\int 2f'(2x)dx=\int \frac{8}{3}x dx$$ $$f(2x)=\frac{4}{3}x^2+c $$ $$ f(x)=\frac{1}{3}x^2+c$$ Where $c$ is a constant. And we clearly have the continuity of $f$ in $0$.

Answer 5

f(2x) = f(x) + x^2

f(x) = f(x/2) + x^2/4

f(x) = f(x/4) + x^2/16 + x^2/4

  = f(x/8) + x^2/64 + x^2/16+ x^2/4

continuing like this

you will get

f(x) = f(x/2^k) + x^2 (1/4+ 1/4^2+......+ 1/4^k)

  = f(x/2^k) + x^2/3(1-(1/4)^k)

for some k ∈ N

k -> infinity

f(x) = f(0)+ x^2/3