9
$\begingroup$

Verify surjectivity of the function $f:\Bbb R \rightarrow \Bbb R$ satisfying: $\forall x \in \Bbb R:f(x)+f(x^2)=x$

I know how to check a function is surjective if we have its explicit formula,but have no idea for this type of question involving a functional equation!

$\endgroup$
  • 2
    $\begingroup$ It might be useful that $f(x)-f(-x)=2x$, by subtracting the relation applied at $x$ to it applied at $-x$. Not sure how to get to surjectivity though... $\endgroup$ – Milo Brandt Jan 1 '18 at 5:19
  • $\begingroup$ Does the problem say anything about continuity of the function? Also, "the function" suggests that there is only one, which seems unlikely. $\endgroup$ – N. S. Jan 1 '18 at 5:31
  • $\begingroup$ @N.S. No, the problem statement says nothing about continuity $\endgroup$ – Hamid Reza Ebrahimi Jan 1 '18 at 6:16
3
$\begingroup$

This is not true unless we assume $f$ is continuous.

Note first that we can use the relation at several points simultaneously to derive the following equation: $$f(x)-(-1)^nf(x^{2 ^n})=\sum_{i=0}^{n-1}(-1)^i\left(f(x^{2^i})+f(x^{2^{i+1}})\right)=\sum_{i=0}^{n-1}(-1)^ix^{2^i}.$$ Now, suppose we define a set $S_x=\{\pm x^{2^n}:n\in\mathbb Z\}$ where $x$ is a positive real number other than $1$. We can use the above equation to determine the value of $f(y)$ from $f(x)$ for every $y\in S_x$ - this follows from observing that $f(x)-f(-x)=f(x)+f(x^2)-f(x)-f(x^2)=x-(-x)=2x$ and using the relation above, noting that it is both necessary and sufficient (and satisfiable). Note that $f(y)$ always depends on $f(x)$ linearly (and non-trivially). As $S_x$ is countably, this implies that, for any $\alpha\in \mathbb R$, we can choose $f$ on $S_x$ such that $\alpha \not\in f[S_x]$.

Note that, since if $y\in S_x$ we have $y^2\in S_x$ and conversely if $y^2\in S_x$ then $y\in S_x$, it actually suffices to show that $f$ satisfies the relation on each $S_x$ individually. However, if we choose $\alpha$ to be neither $f(0)=0$ nor $f(1)=\frac{1}2$ and then fix $f$ to miss this value on every $S_x$ (using the axiom of choice), we find that $f$ is actually not surjective.


If $f$ is continuous, then showing it is surjective amounts to showing that it is neither bounded above nor below.

Choose some $x>1$. Observe that if we divide the first equation in the post by $x^{2^{n-1}}$ and then take a limit as $n$ goes to $\infty$, observing that since $x^{2^n}$ grows faster than any exponential function, we can actually cancel almost the whole sum, we get: $$\lim_{n\rightarrow\infty}\frac{f(x^{2^n})}{x^{2^{n-1}}}=1.$$ Thus, $f$ is unbounded above. However, this then gives that $$\lim_{n\rightarrow\infty}\frac{f(-x^{2^n})}{x^{2^n}}=\lim_{n\rightarrow\infty}\frac{f(x^{2^n})-x^{2^n}}{x^{2^{n}}}= -1.$$ so $f$ is also not bounded below.


One might also note that you can actually write down an equation for $f$ if it is continuous* using the first equation cleverly (basically, sending $n$ to $-\infty$) and noting that $f(1)=1/2$ from $f(1)+f(1^2)=1$. You have to deal with a convergence issue subtly (hence the subtraction in the sum), but otherwise we have, for $x>0$ that: $$f(x)=\frac{1}2-\sum_{i=1}^{\infty}(-1)^i(x^{2^{-i}}-1)$$ Then $f(0)=0$ and $f(-x)=f(x)-2x$.

We can also, for $|x|<1$, sending $n$ to infinity in the first equation, derive a power series around zero: $$f(x)=\sum_{i=1}^{\infty}(-1)^ix^{2^i}.$$ Unfortunately, it is not particularly clear whether the two definitions of $f$ actually agree anywhere, so it's hard to say whether this defines a continuous function.

$\endgroup$
  • $\begingroup$ I feel this problem should have a much more simpler solution $\endgroup$ – Hamid Reza Ebrahimi Jan 1 '18 at 6:55
  • 1
    $\begingroup$ you should also note that in assuming continuity: $x\to\infty:f(x)+f(x^2)=x\implies f(x^2)\lor f(x)\to\infty$ and $x\to\infty:f(-x)+f(x^2)=-x\implies f(x^2)\lor f(-x)\to-\infty$, so if it is continuous function you are done, no need to find $f(x)$ $\endgroup$ – ℋolo Jan 1 '18 at 8:24
  • 1
    $\begingroup$ I am not sure I follow: you say "As $S_x$ is countably, this implies that, for any $\alpha\in \mathbb R$, we can choose $f$ on $S_x$ such that $\alpha \not\in f[S_x]$." Here we are not allowed to define $f$ arbitrarily on $S_x$ we determine one value namely $f(x)$ and the others are automatically defined. How can we be sure we don't hit $\alpha$? $\endgroup$ – clark Jan 1 '18 at 9:02
  • 1
    $\begingroup$ @clark For every $y$ in $S_x$, we have $f(y)=a_{x,y}f(x)+c_{x,y}$ from the given equation, for constants $a_{x,y}$ and $c_{x,y}$. However, $c_{x,y}$ is never zero (it is, in fact, always $1$ or $-1$). In particular, for every $y$, there is exactly one value for $f(x)$ such that $f(y)= \alpha$. However, there are only countably many $y$, thus only countably many values for $f(x)$ such that $\alpha\in f[S_x]$ when we extend. Thus, we just need to choose $f(x)$ not to be in some countable set. $\endgroup$ – Milo Brandt Jan 1 '18 at 16:30

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.