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

About

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 $$\lim_{n\rightarrow \infty}f\left(\frac{1}{2^n}y\right) = f\left( \lim_{n\rightarrow \infty} \frac{1}{2^n}y \right)$$ ?
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  • $\begingroup$ 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
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    $\begingroup$ To get $f(0)=0$ just plug in $x=0$ into your equation $\endgroup$ – GReyes Jan 22 at 20:00
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    $\begingroup$ $f(0)=0$ 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
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    $\begingroup$ 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
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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 $f(x)=\frac13 x$.

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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 $$f(x) =\frac{g(x)}{x}=\frac{x}{3}.$$

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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 $f(x)=\dfrac x3$.

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    $\begingroup$ 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
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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

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

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  • $\begingroup$ 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

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