$f:R\to R$ has the property for $y\in R$, $\sup_{x\in R} [xy-f(x)]$ is finite. Then prove $\lim_{|x|\to\infty} \frac{f(x)}{|x|}=+\infty$ Let $f:\mathbb{R}\to \mathbb{R}$ be function satisfying the property that for all $y\in \mathbb{R}$ the value of the expression $$\sup_{x\in \mathbb{R}} [xy-f(x)]$$ is finite. Define $g(y)=\sup_{x\in \mathbb{R}} [xy-f(x)]$ for $y\in \mathbb{R}$. Then

*

*$f$ must satisfy  $\lim_{|x|\to\infty} \frac{f(x)}{|x|}=+\infty$.

*$g$ is even if $f$ is even.

*$f$ must satisfy  $\lim_{|x|\to\infty} \frac{f(x)}{|x|}=-\infty$.

*$g$ is odd if $f$ is even.


The answer key says $1,2$ both are correct.

I was able to prove that $2$ is correct.
Suppose $f$ is even then $f(-x)=f(x)$. Since $xy-f(x)=(-x)(-y)-f(-x)$
therefore $g(y)=g(-y)$ from the definition of $g$.

But I want help for $1$.
As I'm writing this post, I got this idea:-
Suppose $x>0$, I want to prove $\lim_{x\to\infty} \frac{f(x)}{x}=+\infty$.
Let's use to method of contradiction, i.e. $\exists M>0\,\,\forall N>0\,\,\exists x>N\implies \frac{f(x)}{x}\le M$.
Therefore, $\exists M>0$ and a sequence $x_n\to\infty$ s.t.  $\frac{f(x_n)}{x_n}\le M\implies f(x_n)\le M x_n\implies -f(x_n)\ge -M x_n$.
So, for $y>M$, $g(y)\ge x_n y-f(x_n)\ge x_ny-Mx_n=x_n(y-M)$ so as $x_n\to\infty$, $g(y)\to\infty$ as $n\to\infty$ which is a contradiction.
Is this correct? It seems like I can't do the above for $x<0.$
EDIT:- For x<0, as above assume $\exists M'>0$ and a sequence $x_n'\to-\infty$ s.t.  $\frac{f(x_n')}{x_n'}\le M'$ But since $x_n'<0$ after some $n$, so $f(x_n')\ge M' x_n'\implies -f(x_n)\le -M x_n$.
So,
$g(y)\ge x_n'y-f(x_n')\le x_n'y-M'x_n'$ . Now I'm stuck.
 A: You're quite right! For all $n$ and for any $y>M,$ we would have $$g(y)\geq x_n(y-M),$$ but since $y-M>0$ and $x_n\to\infty,$ this isn't possible.
This doesn't prove it in both directions, but we can fix it. Instead assume that $$\frac{f(x)}{|x|}$$ does not grow without bound as $|x|$ does, so that $$\exists M>0:\forall N>0,\exists x\in\Bbb R:\bigl(|x|>N\bigr)\wedge\left(\frac{f(x)}{|x|}\le M\right).$$ From this, we conclude that there is a sequence of non-zero real numbers $x_n$ such that $|x_n|\to\infty,$ but such that $$\frac{f(x_n)}{|x_n|}\le M$$ for all $n.$ Then $$-f(x_n)\ge -M|x_n|$$ for all $n.$ If infinitely-many of our $x_n$ are positive, we find a contradiction by considering the subsequence of positive $x_n$ just as you did above. Thus, we conclude that there must be only finitely-many $x_n>0,$ and so we may assume without loss of generality that all $x_n<0,$ whence $$-f(x_n)\ge -M|x_n|=Mx_n$$ for all $n.$ Thus, for all $y\in\Bbb R$ and all $n,$ we have $$g(y)\ge x_ny-f(x_n)\ge x_ny+Mx_n=(y+M)x_n.$$ This time, we find the contradiction by choosing $y<-M,$ completing the proof.

I'd also be a little more careful in justifying that $g$ is even if $f$ is even, and actually use the definition of $g$ rather than simply referring to it.
Supposing that $f$ is even, we have that $$g(-y):=\sup\bigl\{x(-y)-f(x):x\in\Bbb R\bigr\}.$$ Given any $t\in\bigl\{x(-y)-f(x):x\in\Bbb R\bigr\},$ there exists $x\in\Bbb R$ such that $t=x(-y)-f(x)$ by definition, but letting $z=-x,$ we see that $t=(-z)(-y)-f(-z)=zy-f(-z),$ so $t=zy-f(z)$ by evenness of $f,$ whence $$\bigl\{x(-y)-f(x):x\in\Bbb R\bigr\}\subseteq\bigl\{zy-f(z):z\in\Bbb R\bigr\}.$$ The reverse inclusion can be shown in a similar fashion, so that $$\bigl\{x(-y)-f(x):x\in\Bbb R\bigr\}=\bigl\{zy-f(z):z\in\Bbb R\bigr\},$$ or $$\bigl\{x(-y)-f(x):x\in\Bbb R\bigr\}=\bigl\{xy-f(x):x\in\Bbb R\bigr\}.$$ Hence, $$g(-y):=\sup\bigl\{x(-y)-f(x):x\in\Bbb R\bigr\}=\sup\bigl\{xy-f(x):x\in\Bbb R\bigr\}=:g(y),$$ as desired.
