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 $f(0)=0$


  • Ok, I know why $f(0) =0$ but why I need informations about "Continuity at a point $0$ " ? It comes to $$\lim_{n\rightarrow \infty}f\left(\frac{1}{2^n}y\right) = f\left( \lim_{n\rightarrow \infty} \frac{1}{2^n}y \right)$$ ?
  • $\begingroup$ Are you sure you meant what you wrote? $2 f(x) = f(x) + x$ just says $f(x) = x$. $\endgroup$ – Robert Israel Jan 22 at 19:57
  • $\begingroup$ Are you sure of your functional equation? $\endgroup$ – lulu Jan 22 at 19:57
  • 3
    $\begingroup$ To get $f(0)=0$ just plug in $x=0$ into your equation $\endgroup$ – GReyes Jan 22 at 20:00
  • 1
    $\begingroup$ $f(0)=0$ because when you set $x=0$ in your (corrected) functional equation you get $2f(0)=f(0)$. $\endgroup$ – Dog_69 Jan 22 at 20:01
  • 1
    $\begingroup$ 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 $f(x)=x/3$ $\endgroup$ – 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 $f(x)=\dfrac x3$.

  • 1
    $\begingroup$ Too all. Am I too restrictive ? Can we conclude assuming only continuity in $0$ and not derivability ? $\endgroup$ – zwim Jan 22 at 20:48
  • $\begingroup$ 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)? $\endgroup$ – LSpice Jan 22 at 21:07
  • $\begingroup$ 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. $\endgroup$ – 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

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


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


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

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

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

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.