# Find all continuous functions in $0$ that $2f(2x) = f(x) + x$

I need to find all functions that they are continuous in zero and $$2f(2x) = f(x) + x$$

I know that there are many examples and that forum but I don't understand one thing in it and I need additional explanation. (Nowhere I see similar problem :( )

## My try

I take $$y= 2x$$ then
$$f(y) = \frac{1}{2}f\left(\frac{1}{2}y\right) + \frac{1}{4}$$ after induction I get: $$f(y) = \frac{1}{2^n}f\left(\frac{1}{2^n}y\right) + y\left(\frac{1}{2^2} + \frac{1}{2^4} + ... + \frac{1}{2^{2n}} \right)$$ I take $$\lim_{n\rightarrow \infty}$$ $$\lim_{n\rightarrow \infty}f(y) = f(y) = \lim_{n\rightarrow \infty} \frac{1}{2^n}f\left(\frac{1}{2^n}y\right) + y\cdot \lim_{n\rightarrow \infty} \left(\frac{1}{2^2} + \frac{1}{2^4} + ... + \frac{1}{2^{2n}} \right)$$ $$f(y) = \lim_{n\rightarrow \infty} \frac{1}{2^n} \cdot f\left( \lim_{n\rightarrow \infty} \frac{1}{2^n}y \right) + \frac{1}{3}y$$

Ok, there I have question - what I should there after? How do I know that $$f(0) = 0$$? I think that it can be related with " continuous functions in $$0$$ " but
function is continous in $$0$$ when $$\lim_{y\rightarrow 0^+}f(y)=f(0)=\lim_{y\rightarrow 0^-}f(y)$$ And I don't see a reason why $$f(0)=0$$

## edit

• Ok, I know why $$f(0) =0$$ but why I need informations about "Continuity at a point $$0$$ " ? It comes to $$\lim_{n\rightarrow \infty}f\left(\frac{1}{2^n}y\right) = f\left( \lim_{n\rightarrow \infty} \frac{1}{2^n}y \right)$$ ?
• Are you sure you meant what you wrote? $2 f(x) = f(x) + x$ just says $f(x) = x$. – Robert Israel Jan 22 at 19:57
• Are you sure of your functional equation? – lulu Jan 22 at 19:57
• To get $f(0)=0$ just plug in $x=0$ into your equation – GReyes Jan 22 at 20:00
• $f(0)=0$ because when you set $x=0$ in your (corrected) functional equation you get $2f(0)=f(0)$. – Dog_69 Jan 22 at 20:01
• You need continuity to conclude that, since $y/2^n\to 0$ as $n\to\infty$, then $f(y/2^n)\to 0$ as well (for any $y$). The only solution is $f(x)=x/3$ – GReyes Jan 22 at 20:35

A powerful method to solve these kinds of problems is to reduce to a simpler equation. In this case we want to eliminate the $$x$$ in the right hand side. Set $$g(x)=f(x)+ax$$, with $$a$$ to be found later. Note that $$f$$ is continuous if and only if $$g$$ is. Then the equality becomes $$2(g(2x)-a(2x))=g(x)-ax+x$$ $$2g(2x)=g(x)+x(1+3a)$$ Therefore setting $$a=-\frac13$$ the equality simplifies to $$g(2x)=\frac12g(x).$$ Now plugging zero gives $$g(0)=0$$. You can now prove by induction that for every $$x$$ $$g\left(\frac{x}{2^n}\right)=2^ng(x).\tag{1}$$ If $$g$$ is not identically zero, say $$g(x_0)\neq 0$$, then we find a contradiction. Indeed by continuity in zero (which is still true for $$g$$) $$g(\frac{x_0}{2^n})$$ should converge to zero, while by $$(1)$$ it does not.

Therefore we conclude that $$g$$ must be identically zero, or equivalently $$f(x)=\frac13 x$$.

Let $$g(x) = xf(x)$$. We obtain $$g(2x) = g(x) +x^2.$$ Since $$\lim\limits_{x\to 0}g(x)=g(0)=0$$, $$\begin{eqnarray} g(x)=g(x) -\lim_{n\to\infty}g(2^{-n-1}x) &=&\sum_{j=0}^\infty g(2^{-j}x)-g(2^{-j-1}x)\\ &=&\sum_{j=0}^\infty 2^{-2j-2}\cdot x^2=\frac{x^2}{3}. \end{eqnarray}$$ This gives $$f(x) =\frac{g(x)}{x}=\frac{x}{3}.$$

Let assume any required regularity for the moment.

Plugging $$x=0$$ give $$2f(0)=f(0)$$ thus $$f(0)=0$$.

Derivating gives $$4f'(2x)=f'(x)+1$$ then $$8f''(2x)=f''(x)$$

So let solve first $$g(2x)=\frac 18 g(x)$$.

$$g(2^n)=a/8^n$$ let assume $$g(x)=\dfrac a{x^3}$$

This would give $$f(x)=\dfrac ax+bx+c$$ continuity in $$0$$ implies $$a=0$$ so $$f(x)=bx+c$$.

$$f(0)=c=0$$ so $$f(x)=bx$$.

$$2f(2x)=4bx=f(x)+x=(b+1)x\iff b=\frac 13$$ so $$f(x)=\dfrac x3$$

Now that we have found what $$f$$ should look like, lets work by substitution.

Set $$f(x)=\dfrac x3g(x)$$ then $$2f(2x)=\dfrac{2x}3g(2x)=\dfrac x3g(x)+x\iff 2g(2x)=g(x)+1$$

Set $$h(x)=g(x)-1$$ then $$2h(2x)+2=h(x)+2\iff 2h(2x)=h(x)$$

In particular $$h(x)=\frac 12h(\frac x2)=\cdots\frac 1{2^n}h(\frac x{2^n})\to 0$$ assuming $$h$$ is bounded in $$0$$.

But we need a little more than continuity of $$f$$ here to conclude $$h=0$$, we need info on $$\dfrac{f(x)}{x}$$ at $$0$$, so since $$f(0)=0$$ we need derivability of $$f$$ in $$0$$.

Assuming this condition then $$f(x)=\dfrac x3$$.

• Too all. Am I too restrictive ? Can we conclude assuming only continuity in $0$ and not derivability ? – zwim Jan 22 at 20:48
• Doesn't the original post already give an argument using only continuity (once the point, which you have proved, about $f$ mapping $0$ to $0$ is resolved)? – LSpice Jan 22 at 21:07
• In fact I should have substituted directly like Del did, it avoids the point f(x)/x. But the method is basically the same. I created an artificial problem. – zwim Jan 22 at 22:07

This recurrence equation is linear then

$$f(x) = f_h(x)+f_p(x)$$

such that

$$a f_h(a x)-f_h(x) = 0\\ a f_p(a x)-f_p(x) = x$$

for the homogeneous equation we assume

$$f_h(x) = \frac Cx$$

and then for the paticular we assume

$$f_p(x) = \frac{C(x)}{x}$$

then

$$a\frac{C(a x)}{a x}-\frac{C(x)}{x} = x$$

or

$$C(a x)-C(x) = x^2$$

for this last recurrence equation we choose

$$C(x) = \frac{x^2}{a^2-1}$$

so the final solution is

$$f(x) = \frac{C}{x}+\frac{x}{a^2-1}$$

in our case $$a = 2$$ then

$$f(x) = \frac Cx+\frac x3$$

and to assure continuity at $$x=0$$ we choose $$C = 0$$ so the final result is

$$f(x) = \frac x3$$

• The function is required to be continuous at $x = 0$, so you'd better have $C = 0$. – LSpice Jan 22 at 23:58
• @LSpice Yes. I will fix that accordingly. Thanks. – Cesareo Jan 23 at 0:01