What is FKMO?

hydo2332 wrote: What is FKMO? Final Korean Mathematical Olympiad.

Could you explain why h is Cauchy? @above: I know the definition. But how do you prove it

My solution: Easy to see that $f$ is surjective and $g$ is injective by plugging in $x=0$. Moreover, changing $x$ for $-x$, we see that $g(-x)=-g(x)$ for all $x \ne 0$ by injectivity. Claim 1: $g$ is unbounded.

Claim 2: $f$ is injective and $g$ is surjective, hence both are bijective.

Now $g(a)=0$ for some $a$, but if $a \ne 0$, then also $g(-a)=0$ contradicting injectivity. Hence $g(0)=0$ and also $f(0)=0$. Now the previous equation $f(-g(y))=g(0)^2-y$ becomes simply that $f$ and $g$ are inverse to each other. Hence $f(x^2-y)=g(x)^2-f(y)$ for all $x,y$ but $y=0$ shows $g(x)^2=f(x^2)$ whence $f(x^2-y)=f(x^2)-f(y)$ and then $f(x+y)=f(x)+f(y)$. On the other hand, $f(x^2)=g(x)^2 \ge 0$ so $f$ is positive for $x$ positive and satisfies Cauchy's equation from where it is well-known that $f$ is linear and then one easily checks that only $f(x)=g(x)=x$ work.

@above Tintarn wrote: Now again from $f(-g(y))=g(0)^2-y$ it follows that $g$ is surjective. Why?

Quasi07 wrote: @above Tintarn wrote: Now again from $f(-g(y))=g(0)^2-y$ it follows that $g$ is surjective. Why? For any $z$, choose $y=g(0)^2-f(-z)$ so that $f(-g(y))=f(-z)$ and hence $g(y)=z$ by injectivity of $f$.

My solution: $f$ is surjective and $g$ is injective. Also, $g(-x)=-g(x)$ when $x\neq 0$. Let $P(x, y)$ denote the equation. $P(x, a): f(x^2-g(a))=g(x)^2-a$. Rewriting this with $t=x^2-g(a)$, we get that for all $t\geq -g(a)$, $f(t)=g(\sqrt{t+g(a)})^2-a$. Putting this back into the original equation, we get: Whenever $x^2\geq g(y)-g(a)$: $g(\sqrt{x^2-g(y)+g(a)})^2-a=g(x)^2-y$ , hence $g(\sqrt{x^2-g(y)+g(a)})^2=g(x)^2+a-y$ when $x^2\geq g(y)-g(a)$ ---(1) Claim: There exists $ M > 0 $ s. t. when $t\geq M ,g(t)>0$.

Choose any four real numbers satisfying $a_1-y_1=a_2-y_2$. Let $x$ be a real number so large such that $x^2\geq g(y_1)-g(a_1)$, $x^2\geq g(y_2)-g(a_2)$, $\sqrt{x^2+g(a_1)-g(y_1)}\geq M$, $\sqrt{x^2+g(a_2)-g(y_2)}\geq M$. Putting this $x$ and $a_i, y_i$ into equation (1), we get $g(\sqrt{x^2+g(a_1)-g(y_1)})^2=g(\sqrt{x^2+g(a_2)-g(y_2)})^2$. Because they are both larger than $M$, the $g$'s are positive and $g(\sqrt{x^2+g(a_1)-g(y_1)})=g(\sqrt{x^2+g(a_2)-g(y_2)})$. Since $g$ is injective, we get $g(a_1)-g(y_1)=g(a_2)-g(y_2)$ for all four numbers such that $a_1-y_1=a_2-y_2$. Hence we have $g(x+y)+g(0)=g(x)+g(y)$ for all $x, y$, and because of the claim we can say $g$ is a linear function. Putting $g(x)=cx+d$ into (1) gives $g(x)=x$, and hence $f(x)=x$ from the original equation.

Nice and not too hard, but kinda lengthy. $f$ has negligible role in the given expression, and we can replace it by the following: $$a^2-g(b)=c^2-g(d) \implies g(a)^2-b=g(c)^2-d$$Call the above $P(a,b,c,d)$. Let $m= \inf\limits_{x \in \mathbb{R}} |g(x)|$ Claim 1: $g$ is strictly increasing. Proof: Assume $g(b) \leq g(d)$ for some $b,d \in \mathbb{R}$. Then for any $a \in \mathbb{R}$, there exists a real number $c$ satisfying $c^2=a^2-g(b)+g(d)$. Now $P(a,b,c,d)$ gives $$g(a)^2-b+d = g(c)^2 \geq m^2$$$$\implies d-b \geq m^2-g(a)^2$$Since this holds for all real numbers $a$, we can take $g(a)^2$ as close to $m^2$ as possible $\implies$ $d-b \geq 0$ $\implies$ $b \leq d$. Taking contrapositive of this, $b>d$ $\implies$ $g(b)>g(d)$, as required. $\blacksquare$ In particular, by Claim 1, $g$ is injective. Claim 2: $g$ is pseudo-odd, i.e., $g(-x)=-g(x)$ for all $x \neq 0$. Proof: $P(x,0,-x,0)$ $\implies$ $g(x)^2=g(-x)^2$ $\implies$ $g(-x)=-g(x)$ by injectivity since $x \neq -x$. $\blacksquare$ Now, for $x>0$, $x>-x$ $\implies$ $g(x) > g(-x) = -g(x)$ $\implies$ $g(x)>0$ $\implies$ $g(x) \geq m$. So, $g(x) \leq -m$ for $x <0$. If $|g(0)|>m$, then there exists a real number $x$ with $|g(0)|>|g(x)| \geq m$. We can take $x>0$ due to pseudo-oddness. Now, we either have $g(0)>g(x)$, or $g(0)<-g(x)=g(-x)$, both contradicting Claim 1. Therefore $g(0)= \pm m$. Let $R$ be the range of $g$. Claim 3: $|k|>m$ $\implies$ $k \in R$. Proof: By Claim 1, we have $g(k^2-m^2)>g(0)$. Therefore there exists a real number $c$ such that $c^2-g(k^2-m^2)=0^2-g(0)$. $$\implies g(c)^2=k^2-m^2+g(0)^2=k^2$$Also, $c \neq 0$; otherwise, $m^2=g(c)^2=k^2>m^2$, contradiction. Therefore $k=g(c)$ or $k=g(-c)$ depending on sign of $k$ ny pseudo-oddness, as required. $\blacksquare$ Now, assume $g(0)=m$ (similar method will work when $g(0)=-m$). Therefore $[m, +\infty) \subseteq R$ by Claim 3 (in particular, $[m, +\infty) \subseteq g(\mathbb{R}_{\geq 0})$). Also, $g(x)>m$ for $x>0$ by injectivity. For any $x,y > 0$, $g(x)+g(y)-m > m$ $\implies$ $g(x)+g(y)=g(z)+m=g(z)+g(0)$ for some $z > 0$. Now choose any $a,c \in \mathbb{R}$ such that $$a^2-c^2= g(x)+g(y)=g(z)+g(0)$$Now using pseudo-oddness and comparing $P(a,x,c,-y)$ and $P(a,0,c,-z)$, we get $$x+y=g(a)^2-g(c)^2=z$$Therefore $g(x)+g(y)=g(x+y)+m$ for all $x,y>0$. So the function $g(x)-m$ satisfies Cauchy's functional equation on $\mathbb{R}^+$, and is bounded below by $0$; therefore $g(x)=m+px$ for some constant $p$. $P(a,b,c,d)$ transforms into $$a^2-m-pb=c^2-m-pd \implies (m+pa)^2-b=(m+pc)^2-d$$for all $a,b,c,d>0$. Therefore $$a^2-c^2=pb-pd=p((m+pa)^2-(m+pc)^2)=2mp^2(a-c)+p^3(a^2-c^2)$$for all $a,c>0$. Now, comparing coefficients, we get $p=1$ and $m=0$ $\implies$ $g(x)=x$ for $x>0$ and $g(0)=0$. So by pseudo-oddness, $g(x)=x$ for all $x$. Now bringing back $f$, $f(x^2-y)=x^2-y$ for all $x,y \in \mathbb{R}$. Now applying the recently proven deep result known as "triviality", we can find $f$ as well: $$\boxed{f(x)=g(x)=x \ \ \forall x \in \mathbb{R}}$$$\blacksquare$

Ultra nice problem The only solution is $f(x)=x$ and $g(x)=x$. Obviously we must have that $f$ is surjective, while $g$ must be injective. Then notice that by symmetry of $x^2=(-x)^2$, that we have that $f(x^2-g(y))=f((-x)^2-g(y))$, i.e. $g(x)^2=g(-x)^2$. Since $g$ is injective we must have that $g(-x)=-g(x)$ for all non-zero real numbers. This implies that $f(x^2+g(t))=g(x)^2+t$. Let's say that for some $x_1,x_2 > 0 $, we have that $f(x_1-g(y))=f(x_2-g(y))$, i.e. $g(\sqrt{x_1})^2=g(\sqrt{x_2})^2$, this implies that $g(\sqrt{x_1}) = g(\sqrt{x_2}) \implies x_1=x_2$,since if it were otherwise we would have that $g(\sqrt{x_1}) = -g(\sqrt{x_2}) = g(-\sqrt{x_2} \implies \sqrt{x_1} = -\sqrt{x_2}$, which is nonsense, since $\sqrt{x_1} > 0$. Thus we have that in an open interval which is either $[-\text{max} \; g(x),+\infty)$, or $[\text{min} \; g(x),+\infty)$. Now let's say that $|g(x)| \leq M$, for all $x$. Then we must have that for any $y_1,y_2$, there exist $x_1$ and $x_2$ such that $x_1^2-x_2^2=g(y_1)-g(y_2) \implies x_1^2-g(y_1)=x_2^2-g(y_2)$, then when we put this into our asseretion we get that $g(x_1)^2-y_1=g(x_2)^2-y_2 \implies g(x_1)^2-g(x_2)^2=y_1^2-y_2^2 \in [-2M,2M]$. But since we had that $y_1$ and $y_2$ were arbitrary, we can easily break the boundry of $M$, i.e. $g$ is unbounded. This implies that $f$ is injective for all real numbers, and since it is also surjective that implies that $f$ is a bijection. Let's set $y=g(x)^2$, into our assertion then we have that $f(x^2-g(g(x)^2))=0$, since $f$ is a bijection there exists some unique $t$ for which $f(t)=0$, this implies that: $$g(g(x)^2)=x^2-t$$since $g$ is odd we have that: $$g(-g(x)^2)=t-x^2$$i.e. from these relations we have that $g$ is also surjective, i.e. $g$ is a bijection. Since $g$ is a bijection there exists some unique $k$ for which $g(k)=0 \implies g(-k)=0 \implies k=-k \implies k=0$. Setting $y=0$, we have that $f(x^2)=g(x)^2$, this gives us $f(0)=0$. Setting $x=0$, we have that $f(-g(y))=-y \implies f(g(-y))=-y$, i.e. we get that $g \equiv f^{-1}$. This implies that $g(f(x))=x$. Now let's set $y=f(x^2)$, to get that $f(x^2)=x^2$, i.e. for any $x \in \mathbb{R}^{+}$ we have that $f(x)=x$. This gives us that $x=g(f(x))=g(x)$, where $x$ is a positive real number. This implies that $g(-x)=-x \implies f(-x)=f(g(-x))=-x$ for any positive real $x$. Thus we get that the only pair that satisifies is $(f(x),g(x))=(x,x)$

The answer is $f(x)=g(x)=x$ only, which clearly works. By fixing $x$ and varying $y$ it follows that $f$ is surjective and $g$ is injective. Then by swapping $x$ with $-x$, we find that $g$ is "pseudo-odd", so $f(x)=-f(-x)$ when $x \neq 0$. By picking $y$ such that $|y|$ is large, $g(y)$ is positive, and letting $x=\sqrt{g(y)}$ implies that $g(x)^2$ is unbounded, since it differs from $y$ by a constant $f(0)$. Thus $|g(x)|$ is unbounded so $g(x)$ is unbounded in both directions by pseudo-oddness. Plug in some $y$ such that $-g(y)$ is very negative and vary $x>0$: this establishes the injectivity of $f$, using the injectivity and pseudo-oddness of $g$. Hence $f$ is a bijection. By fixing $y$ such that $-g(y)$ is very negative and varying $x$ implies that $g^2$ is surjective, so by pseudo-oddness $g$ is almost surjective, with the sole exception being that $-g(0)$ might not be in the range of $g$. However, I actually claim that $g(0)=0$: indeed, if not we have some $x \neq 0$ such that $g(x)=0$ by "almost-surjectivity" and $g(-x)=0$ from pseudo-oddness: contradiction to the injectivity of $g$. Hence $g$ is indeed surjective, hence also a bijection. Let $h=f^{-1}$, which is a bijection, and rewrite the equation as $x^2-g(y)=h(g(x)^2-y)$. Plugging in $x=0$, we have $-g(y)=h(-y) \implies g(y)=h(y)$, since $g$ is odd (pseudo-odd and $g(0)=0$ implies odd). Thus the equation now becomes $x^2-g(y)=g(g(x)^2-y)$. By oddness it follows that $x^2+g(-y)=g(g(x)^2-y)$, and since $g$ is a bijection it follows that $g(a)-g(b)=g(a+k)-g(b+k)$ for any $a,b,k$; in particular by letting $k=-b$ we extract $g(a)=g(b)+g(a-b)$, so $g$ is additive. On the other hand, by varying $x$ in our original and fixing $y$ it follows (since $g$ is a bijection) that $g$ is bounded below on $[-y,\infty)$, hence $g$ is of the form $x \mapsto rx$. It is now trivial to check that $g(x)=x$ is the only soltuion. $\blacksquare$ Remark: My original solution didn't involve $g=h$ sine I didn't realize this was true (nor that $g(0)=0$). Instead, one notes that $g(x)-g(y)=h(k+x)-h(k+y)$ for any $k \geq 0$, which you can use to prove $h(a)+h(b)=h(a+b)+h(0)$ and have something similar for $g$. Then since $h$ is bounded below as before and we check linear solutions.

Answer. $f(x)=x, g(x)=x$ $\forall x \in \mathbb{R}$ Solution. Clearly, $f$ is surjective and $g$ is injective. $f(x^2-g(y))=g(x)^2-y=f((-x)^2-g(y))=g(-x)^2-y \implies -g(x)=g(-x) \ \forall x \in \mathbb{R} \setminus \{0\}$ because of injectivity of $g$. $f(x^2-g(0))=g(x)^2$. Let $z := x^2-g(0)$. Then, $f(z)=g\left(\sqrt{z+g(0)}\right)^2 \ge 0 \implies f(z) \ge 0 \ \forall z \ge -g(0)$. Notice that $x^2 \ge g(y)-g(0)$ implies $x^2-g(y) \ge -g(0) \implies f(x^2-g(y)) \ge 0 \implies g(x)^2-y \ge 0$. To sum up, $x^2 \ge g(y) - g(0) \implies g(x)^2 \ge y$. Lemma 1. $g$ is unbounded. (Note: Since $g$ is an "almost odd" function, it is bounded from above if and only if it is bounded from below) Proof. FTSOC assume that there exists $c \in \mathbb{R}_{+}$ for which $\vert g(x) \vert \le c \ \forall x \in \mathbb{R}$. Choose $y > c^2$ and $x > \sqrt{\vert c-g(0) \vert}$. Then, $x^2 > c - g(0) \ge g(y)-g(0) \implies g(x)^2 \ge y > c^2 \implies \vert g(x) \vert > c$, contradiction. Therefore, $g$ is unbounded. $\ \blacksquare$ Lemma 2. $f, g$ are both bijective. Proof. $\bullet$ $f$ is injective. Assume that $f(a) = f(b)$ for some real numbers $a$ and $b$. Because of $g$ being unbounded, we can choose sufficiently large $y$ for which $g(y)+a>0, g(y)+b>0$. Let $x, z \in \mathbb{R}_{+}$ be the numbers satisfying $g(y)+a=x^2$ and $g(y)+b=z^2$. Then, $g(x)^2-y = f(x^2-g(y)) = f(a) = f(b) = f(z^2 - g(y)) = g(z)^2-y$, therefore, $g(x) = g(z)$ or $g(x)=-g(z)=g(-z)$ because $z \neq 0$. In the first case, $x=z$ because of the injectivity of $g$ and thus, $x=z \implies x^2=z^2 \implies g(y)+a = g(y)+b \implies a=b$. In the second case, $x = -z$ which is impossible since $x, z >0$. Therefore, $f$ is injective. $\bullet$ $g$ is surjective. Clearly, by fixing $x$ and running $y$ over the set $\mathbb{R}$, $f(x^2-g(y))$ runs over all elements of $\mathbb{R}$ and since $f$ is bijective, $x^2-g(y)$ must also run over all elements of $\mathbb{R}$. Thus, $x^2-g(y)$ is surjective and consequently, $g$ is surjective. We got that $f$ and $g$ are both bijective. $\ \blacksquare$ Lemma 3. $f(x) = kx$ for all real $x$ and some non-zero real constant $k$. Proof. Choose $\alpha \in \mathbb{R}$ for which $g(\alpha)=0.$ Assume that $\alpha \neq 0$. Then $f(x^2)=f(x^2-g(\alpha)) = g(x)^2 - \alpha$ and $f(x^2)=f(x^2+g(\alpha)) = f(x^2 - g(-\alpha)) = g(x)^2 + \alpha \implies \alpha=0$, contradiction. Thus, $\alpha = 0 \implies g(0) = 0$. Plugging $x = y = 0$ into the original equation gives us $f(0) = 0$. Notice that $f(x^2)=g(x)^2 \implies f(x) \ge 0 \ \forall x \ge 0$. Since $g(0)=0$ and $g(-x)=-g(x) \ \forall x \neq 0$, $g$ is an odd function. \[ f(x^2-g(y)) = g(x)^2-y \]\[f(x^2+g(y))=g(x)^2+y \]Let $A := x^2-g(y)$ and $B:= x^2+g(y)$. Since $g$ is surjective, both $A$ and $B$ can attain any real value. Therefore, $f(A)+f(B)=2g(x)^2=2f(x^2)=2f\left(\frac{A+B}{2}\right)$ for all real numbers $A$ and $B$ for which $A+B \ge 0$. Choose $B=0$ and $A \ge 0$. Then, $f(A) = 2f\left( \frac{A}{2}\right) \implies f(A)+f(B) = f(A+B)$ for all real numbers $A$ and $B$ satisfying $A+B \ge 0. \ \ \ \ \ (1)$. Let $x, y$ be any real numbers satisfying $x+y \le 0$. Choose sufficiently large $z \in \mathbb{R}$ such that $x+z \ge 0, x+y+z \ge 0$. Then, $f(x+y+z) = f(x+y) + f(z)$ and $f(x+y+z) = f(x+z) + f(y) = f(x) + f(z) + f(y) \implies f(x+y) = f(x) + f(y). \ \ \ \ \ \ \ \ \ \ (2)$ $(1)$ and $(2)$ imply that $f(x+y) = f(x)+f(y) \ \forall x,y \in \mathbb{R}$. Now we prove that $f$ is strictly increasing. Assume that $a > b$. Then, there exists $c > 0$ for which $a = b + c \implies f(a) = f(b)+f(c) > f(b)$. Now clearly $f$ is a Cauchy function and thus $f(x)=kx$ for some $k \in \mathbb{R}$. Obviously, $k$ can't be equal to $0$. $\ \blacksquare$ Plugging $x=0, y = -y$ into the original equation implies $f(g(y))=y \implies g(y)=\frac{y}{k} \ \forall y \in \mathbb{R}$. Then, $f(x^2-g(y))=k(x^2-\frac{y}{k}) = kx^2-y = g(x)^2-y = \frac{x^2}{k^2} - y \implies k^3 = 1 \implies k = 1$. Thus, $f(x)=g(x)=x$ and we are done. $\ \square$