Functional eq

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#1 h Mar 4, 2016, 5:22 AM • 2 Y

Y by anantmudgal09, Adventure10

Find all functions $f\colon \mathbb{R} \to \mathbb{R}$
$f(x^2 + y + f(y)) = f(x)^2 + 2y$ for all $x,y \in\mathbb{R}$

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SohamSchwarz119

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#2 h Mar 4, 2016, 7:38 AM • 3 Y

Y by CmIMaTh, Adventure10, Mango247

Solution:

If we fix $x$ then the right-hand side is surjective on $\mathbb{R}$

therefore $f$ is surjective. Also if $y_1 + f(y_1) = y_2 + f(y_2)$ then writing the condition for some $x$ and $y$ = $y_1$ , $y_2$ the get $2y_1= 2y_2$ thus

$x+f(x)$ is one-one.
Thus for some $c$ we have $f(c) = 0$ . This implies $f(c^2+y+f(y))=2y$

Pick two fixed $a, b$ . Set $c = b+f(b)-a-f(a) > 0$ . If $x > a + f(a)$ then there is an $u$ such that $x = u^2 + a + f(a)$

, $x + d =u^2+b+f(b)$

So, we have: $f(x)=2a+f^2(x)$ and $f(x+c)=2b+f^2(x)$

we conclude that $f(x + c) - f(x) = 2(a - b)$ for all sufficiently big $x$ .

This means that $f(x + c) = f(x) + d$ for $d = 2(b - a)$ and $c + d =3(b - a) + f(b) - f(a)$ and from here $f(x + nc) = f(x) + nd$ for all sufficiently big $x$

and any fixed natural $n$ . If $d < 0$ then $f(x + nc) < 0$ for all sufficiently big n. However if $x+nc > f(0)$ then $x+nc = u^2+f(0)$ and applying the

condition for $u$ and $0$ we get $f(x + nc) = f^2(u)>0$ (contradiction!!)

So $d > 0$ . Then $f(x+nc)+x+nc = f(x)+x+n(c+d)$ .

This means $f(x) + x$ takes arbitrarily large values. Take now a $y$ . We then get $f((x + c)^2+y+f(y))=2y+(f(x)+d)^2$ for $x\ge x_0$

now that we have $f(x+u) = f(x)$ for all sufficiently big $x$ when $u > 0$ .

Then we get $f((x + u)^2+y+f(y)=f(x^2+y+f(y))$

we can choose $x$ such that $2xu + u^2=lc$ , $l>0$ and then choosing $y$ such that $y + f(y)$ is big enough we have $f((x + u)^2+y+f(y))=f(x^2+2xu+u^2+y+f(y))=f(x^2+y+f(y))+ld$ (contradiction!!)

Consider now $g(x) = x + f(x)$ . Take a fixed $s$ and $x \neq y$ with $g(x +s) - g(x) \le g(y + s) - g(y)$ . Then we have $f(t + g(x + s) - g(x)) = f(t) + 2s$ ,

$f(t + g(y + s) - g(y)) = f(t) + 2s$ for all $t \ge t_0$ and if we set $z = t+g(x+s)-g(s))$ , $v = (g(y+s)-g(y))-(g(x+s)-g(x))$ then the identity turns to $f(z + v) = f(z)$ for all $z \ge 0$ . As $v \ge 0$ we must have $v = 0$ as we have proven such a relation is impossible for $v \neq 0$ . Hence,

$g(x + s) - g(x) = g(y + s) - g(y)$ . As $x, y$ were chose arbitrarily we conclude that $g(x+s)-g(x)$ is independent of $x$ thus $g(x+s)-g(x) =g(s) - g(0)$ so $g(x + s) + g(0) = g(x) + g(s)$ thus $h(x) = g(x) - g(0)$ is an additive function, hence so is $f(x) - g(0) = f(x) - f(0)$ . We then get

$f(x^2+g(y))=f(x^2)+f(g(y))-f(0) =2y+f^2(x)$ .Thus $f(x^2)-f^2(x)=2y-f(g(y))$

. If we now fix $y$ then we get $f(x^2)=f^2(x)+e$ for fixed $e$ hence $f(x) \ge -e$ for $x \ge 0$ . Hence $f(x) - f(0)$ is bounded below..it is well known $f(x) - f(0) = rx$ for some $r$ hence $f = rx + s$ is linear. The condition $f(x^2)-f^2(x) = 2y-f(g(y))$ now implies $rx^2+s-(rx+s)^2=2y-r((r+1)y+s)-s$ or $r(1-r)x^2-2rsx-s^2=(2-r(r+1))y-(r+1)s$

which is possible only when $r(1 - r) = 2rs = 2 - r(r + 1) = 0$ . Then $r = 0$ or $r = 1$ . If $r = 0$ then $2 -r(r + 1) \neq 0$ . So $r = 1$ and then $s = 0$ . So $f$ is the identity function..

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SohamSchwarz119

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#3 h Mar 4, 2016, 8:10 AM • 2 Y

Y by Adventure10, Mango247

Please tell whether I am correct

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#4 h Mar 4, 2016, 11:13 AM • 2 Y

Y by Adventure10, Mango247

Let $P(x,y)$ be the assertion $f(x^2+y+f(y))=f(x)^2+2y$
$f$ is obviously surjective.

Let there exist a $k \in \mathbb{R}$ such that $f(k)=0$ .

$P(-k,0)$ and $P(k,0)$
$\implies f(-k)^2=f(k)^2$ $\implies f(-k)=0$
$P(0,k) \implies f(k)=f(0)^2+k$ $\implies k \leq 0$
But similarly
$P(0,-k) \implies -k \leq 0$
Hence $k$ can only be zero.
Due to surjectivity of $f$ , $\boxed{f(0)=0}$

$P(x,0) \implies \boxed{f(x^2)=f(x)^2}$
Thus is $t > 0$ , $f(t)=f(\sqrt{t})^2 > 0$
That is $t > 0 \implies f(t) > 0$

$P(0,y) \implies \boxed{f(y+f(y))=2y}$
Let there exist some $b$ such that $f(b)<0$
$P(\sqrt{-f(b)},b) \implies f(b) = f(\sqrt{-f(b)})^2+2b > 2b$
Let $z <0$
$f(z+f(z))=2z <0$ Using the above result,
$2z=f(z+f(z)) > 2z+2f(z)$

Hence we get if $z < 0 \implies f(z) <0$

From $P(x,0)-P(-x,0)$ and the result of sign (positive or negative) of $f(x)$ depending on $x$ ),
$f(x)+f(-x)=0$
Observe that
$P(x,y) \implies f(x^2+y+f(y))=f(x^2)+2y$ $P(x,y+f(y)) \implies f(x^2+y+f(y)+f(y+f(y))) = f(x^2)+2(y+f(y))$ $\implies f(x^2+2y+y+f(y))=f(x^2)+2y+2f(y)$ for $x^2+2y \ge 0$
$\implies f(x^2+2y+y+f(y))=f(x^2+2y)+2y$
For $x^2+2y \ge 0$
Hence we get $f(x^2+2y)=f(x^2)+2f(y)$
Setting $y=-\frac{x^2}{2}$ above, we get $f(x^2)=-2f(\frac{-x^2}{2})$
This implies $f(2m)=2f(m)$

Thus $f(x^2+2y)=f(x^2)+f(2y)$

But again because of the fact $xf(x) \ge 0$ (after some case making I guess),
we get $f$ is additive that is $\boxed{f(x+y)=f(x)+f(y)}$

Also obviously by the above condition and the fact that $xf(x)>0$ , the function is monotonous...

Hence by Cauchy equation, the function is linear.
Putting value $\boxed{f(x)=x}$ is the only function satisfying the condition and hence we are done.

I will think about an alternate finish though.

This post has been edited 6 times. Last edited by utkarshgupta, Mar 4, 2016, 11:55 AM

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#5 h Mar 4, 2016, 11:39 AM • 1 Y

Y by Adventure10

This problem has been posted many times on this forum. See e.g. here.

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