6
$\begingroup$

find all continuous functions $f:\mathbb{R}^n \rightarrow \mathbb{R}$ satisfying \begin{equation*} f(x+y)+f(x-y)=2f(x)+2f(y)~\forall x,y \in \mathbb{R}^n. \end{equation*} My attempt: I manage to show that for any $q \in \mathbb{Q}$, $f(qx)=q^2f(x)$ for all $x \in \mathbb{R}^n$. I have a feeling that the answer is $f(x)=A\| x \|^2$, but I'm unable to prove it.

Can anyone give some hint?

$\endgroup$
13
  • 3
    $\begingroup$ Then by continuity, $f(rx) = r^2f(x)$ for all $r \in \mathbb R$. $\endgroup$ – user99914 May 25 '15 at 6:58
  • 1
    $\begingroup$ For all $r\in \mathbb R$, let $q_n \in \mathbb Q$ and $q_n \to r$. Then $ r^2 f(x) = \lim q_n^2 f(x) = \lim f(q_n x) = f (\lim q_n x) = f(rx)$. $\endgroup$ – user99914 May 25 '15 at 7:12
  • 2
    $\begingroup$ @John: I don't know yet, or I would have posted it as a solution. $\endgroup$ – Christian Blatter May 25 '15 at 8:48
  • 1
    $\begingroup$ It might help to check out the wiki page on the polarization identity: en.wikipedia.org/wiki/Polarization_identity $\endgroup$ – muaddib May 25 '15 at 9:05
  • 1
    $\begingroup$ It seems that this is sometimes called quadratic functional equation. See, for example, Functional Equations and Inequalities with Applications By Palaniappan Kannappan, p.220 $\endgroup$ – Martin Sleziak May 28 '15 at 14:56
3
+50
$\begingroup$

YES, $f$ must be a quadratic homogeneous polynomial (QHP in abbreviated notation) as you expected.

We have a continuous function $f:\mathbb{R}^n \rightarrow \mathbb{R}$ satisfying $$ f(x+y)+f(x-y)=2f(x)+2f(y)~\forall x,y \in \mathbb{R}^n \tag{1} $$

Let $a,b\in{\mathbb R}^n$. For $k\in{\mathbb N}$, define $\phi_{a,b}(k)=f(a+kb)$. Taking $x=a+(k+1)b$ and $y=b$ in (1) above, we see that $\phi_{a,b}$ satisfies a second-order linear recurrence formula :

$$ \phi_{a,b}(k+2)-2\phi_{a,b}(k+1)+\phi_{a,b}(k)=2f(b) \tag{2} $$

It follows easily from (2), by induction on $k\in{\mathbb N}$ that

$$ \phi_{a,b}(k)=k\phi_{a,b}(1)-(k-1)\phi_{a,b}(0)+k(k-1)f(b) \tag{3} $$

Note that (3) can be rewritten as (putting $\mu=k$)

$$ f(a+\mu b)=f(a)+\mu(f(a+b)-f(a)-f(b))+\mu^2f(b) \ (\mu\in{\mathbb N}) \tag{3'} $$

Now, let $\lambda,\mu\in{\mathbb N}$. We have \begin{equation} \begin{split} f(\lambda a+\mu b) &= f(\lambda a)+\mu\Bigg(f(\lambda a+b)-f(\lambda a)-f(b)\Bigg)+\mu^2f(b) \\ &= \lambda ^2 f(a)+\mu\Bigg(\Bigg[f(b+\lambda a)\Bigg]-\lambda^2 f(a)-f(b)\Bigg)+\mu^2f(b) \\ &= \lambda ^2 f(a)+\mu\Bigg(\Bigg[f(b)+\lambda(f(a+b)-f(a)-f(b))+\lambda^2 f(a)\Bigg] -\lambda^2 f(a)-f(b)\Bigg)+\mu^2f(b) \\ &= \lambda ^2 f(a)+\lambda\mu\Bigg(f(a+b)-f(a)-f(b)\Bigg)+\mu^2f(b) \ \ \ \ \ \ \ \ \ \ \ \ \ (4) \\ \end{split} \end{equation}

Replacing $a$ or $b$ with their opposites, we see that (4) still holds when $\lambda$ or $\mu$ is negative. By homogeneity, (4) still holds when $\lambda,\mu\in{\mathbb Q}$. By continuity, (4) still holds when $\lambda,\mu\in{\mathbb R}$. So the restriction of $f$ to any two-dimensional subspace is a QHP.

As a special case of (4), we have for $x,a\in{\mathbb R}^n$ and $\lambda \in {\mathbb R}$,

$$ f(x+\lambda a)=f(x)+\lambda (f(x+a)-f(x)-f(a)) +\lambda^2 f(a) \tag{5} $$

Viewing $x+\lambda a+\lambda a'$ as $(x+\lambda a)+\lambda' a'$ and using (5) repeatedly, we obtain for $x,a,a'\in{\mathbb R}^n$ and $\lambda,\lambda' \in {\mathbb R}$

$$ \begin{array}{lcl} f(x+\lambda a+\lambda' a') &=& f(x)+(\lambda+\lambda')(f(x+a)-f(x)-f(a))+\\ & & \lambda \lambda' (f(x+a+a')-f(x+a)-f(x+a')+f(x)) +\\ & & \lambda^2f(a)+{\lambda'}^2f(a') \tag{6} \end{array} $$

Replacing $x$ with $\lambda''a''$ in (6) above and using (5) repeatedly one more time, we obtain for $a,a',a''\in{\mathbb R}^n$ and $\lambda,\lambda',\lambda'' \in {\mathbb R}$,

$$ \begin{array}{lcl} f(\lambda a+\lambda' a'+\lambda''a'')&=& \lambda \lambda' (f(a+a')-f(a)-f(a')) + \\ & &\lambda \lambda'' (f(a+a'')-f(a)-f(a'')) + \\ & & \lambda' \lambda'' (f(a'+a'')-f(a')-f(a'')) + \\ & & \lambda^2f(a)+{\lambda'}^2f(a')+{\lambda''}^2f(a'') + \\ & & \lambda\lambda'\lambda''(f(a+a'+a'')-f(a+a')-f(a+a'')-f(a'+a'')+f(a)+f(a')+f(a'')) \\\tag{7} \end{array} $$

Note that when $\lambda=\lambda'=\lambda''=t$, the LHS reduces to $t^2f(a+a'+a'')$ which has no $t^3$ term. So in (7) above the $\lambda\lambda'\lambda''$-coefficient must be zero. Then (7) shows that the restriction of $f$ to any three-dimensional subspace is a QHP. If we put $B(a,b)=f(a+b)-f(a)-f(b)$, it follows that for any $a$, $B(a,.)$ is linear on any two-dimensional subspace, and hence linear everywhere. So $B$ is bilinear, which concludes the proof.

$\endgroup$
0
0
$\begingroup$

I'm going to do the case where $n=2$, I'll leave the generalisation to you :) First, by continuity, for all $i\in\{1,2\}$ and $r\in\mathbb{R}$, $f(r{e_i})=r^2f(e_i)$. Now, denote $f(e_i)=A_i$.

Then, for every $r_1,r_2\in\mathbb{R}$, you know by the given identity that: $$f(r_1,r_2)=2f(r_1,0)+2f(0,-r_2)-f(r_1,-r_2)=2r_1^2A_1+2r_2^2A_2-f(r_1,-r_2)$$ So we just need to know what $f(r_1,-r_2)$ is. Now: $$f(x+y)+f((\sqrt{2}-1)x-(\sqrt{2}+1)y)=2f(\frac{x-y}{\sqrt{2}})+2f((1-\frac{1}{\sqrt{2}})x+(1+\frac{1}{\sqrt{2}})y)=f(x-y)+2f((1-\frac{1}{\sqrt{2}})x+(1+\frac{1}{\sqrt{2}})y)$$ So: $$f(x+y)-f(x-y)=2f((1-\frac{1}{\sqrt{2}})x+(1+\frac{1}{\sqrt{2}})y)-f((\sqrt{2}-1)x-(\sqrt{2}+1)y)=f((\sqrt{2}-1)x+(\sqrt{2}+1)y)-f((\sqrt{2}-1)x-(\sqrt{2}+1)y)=0$$ Therefore: $$f(r_1,r_2)=f(r_1,-r_2)$$ And so, going back to our previous result:: $$f(r_1,r_2)=r_1^2A_1+r_2^2A_2$$

EDIT: It seems there's something wrong with my argument, as $f(r_1,r_2)=r_1r_2$ also satisfies the identity supplied. I'm not sure right now what it is, though, as of now...

$\endgroup$
4
  • 1
    $\begingroup$ There's something wrong in your argument as the outcome should also contain something like $f(r_1, r_2) = r_1r_2$. $\endgroup$ – user99914 May 25 '15 at 8:09
  • 1
    $\begingroup$ Hmm, you're right. Not sure what it is though... $\endgroup$ – Roi Blumberg May 25 '15 at 8:20
  • $\begingroup$ Why $f((\sqrt{2}-1)x+(\sqrt{2}+1)y)-f((\sqrt{2}-1)x-(\sqrt{2}+1)y)=0$? $\endgroup$ – Idonknow May 25 '15 at 13:00
  • $\begingroup$ Whoops, my mistake! $\endgroup$ – Roi Blumberg May 25 '15 at 19:25

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.