A beautiful FE

For discussing Olympiad Level Algebra (and Inequality) problems
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Thanic Nur Samin
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A beautiful FE

Unread post by Thanic Nur Samin » Thu Feb 02, 2017 7:38 pm

Find all continuous functions $f:\mathbb{R}^{\ge 0}\rightarrow \mathbb{R}^{\ge 0}$ so that for any $x,y$ in the domain, we have:

\[f(xf(y))+f(f(y))=f(x)f(y)+2\]
Hammer with tact.

Because destroying everything mindlessly isn't cool enough.

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Thanic Nur Samin
Posts:176
Joined:Sun Dec 01, 2013 11:02 am

Re: A beautiful FE

Unread post by Thanic Nur Samin » Thu Feb 02, 2017 7:54 pm

Let $P(x,y)$ denote the FE, and also assume that $c=f(1)$.

Now, $P(1,1)\Rightarrow f(c)=\dfrac{c^2}{2}+1$

$P(c,x)\Rightarrow f(cf(x))+f(f(x))=(\dfrac{c^2}{2}+1)f(x)+2$
$P(f(x),1)\Rightarrow f(c(f(x))+\dfrac{c^2}{2}+1=cf(f(x))+2$

From which we get $f(f(x))=\dfrac{c^2/2+1}{c+1}f(x)+\dfrac{c^2/2+1}{c+1}$.

Also, $P(1,x)\Rightarrow f(f(x))=\dfrac{c}{2}f(x)+1$.

Note that if $f$ is constant we run into a contradiction, so there must be at least two different values of $f$. So we get $\dfrac{c^2/2+1}{c+1}=1\Rightarrow c=2$, and thus $f(1)=2$ and $f(2)=3$. Also, $f(f(x))=f(x)+1$. Applying induction, we get $f(n)=n+1$ for all positive integers $n$.

Now, if $y=q-1$ is an integer, then the functional equation implies $f(qx)+(q-1)=qf(x)$, where plugging in $x=\dfrac{p}{q}$ we get $f(p/q)=p/q+1$.

From continuity, we deduce that $f(0)=1$ and $f(x)=x+1$ for irrational numbers.

So, the solution is $\boxed{f(x)=x+1}$.
Hammer with tact.

Because destroying everything mindlessly isn't cool enough.

rah4927
Posts:110
Joined:Sat Feb 07, 2015 9:47 pm

Re: A beautiful FE

Unread post by rah4927 » Fri Feb 03, 2017 5:30 pm

We will show that $f(x)=x+1$. Check to see that this indeed verifies the original equation.

$P(0,x)$ implies $f(f(x))=f(0)((f(x)-1)+2$ whilst $P(1,x)$ implies $f(f(x))=\dfrac{f(x)f(1)+2}{2}$. Combining the two gives us $f(x)=\dfrac{2f(0)-2}{2f(0)-f(1)}$ implying $f$ is a constant, which is readily false. So we must have a zero denominator, giving $2f(0)=f(1)$.

$P(0,0)$ yields $f(0)+f(f(0))=f(0)^2+2$ but we know $f(f(0)=\dfrac{f(0)f(1)+2}{2}=\dfrac{2f(0)^2+2}{2}=f(0)^2+1$. Substituting $f(f(0))$ in the former equation results in $f(0)=1$, so $f(1)=2$. We now see from $P(1,x)$ that $f(f(x))=f(x)+1$. This easily gives us $f(x)=x+1$ for all positive integers by induction. Now using $P(\frac{m}{n},n-1)$ for positive integers $m,n$ gives us $f(\frac{m}{n})=\dfrac{m}{n}+1$, finishing the proof for rationals. Now note that since $f(x)$ is a polynomial for all rationals $x$ and also continuous over the reals, we must have $f(x)=x+1$ over all reals (think of this as approaching an irrational from two sides using only rationals).

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