Find all functions from real numbers to real numbers that satisfy the following equation:
$$f(xy) =
\left\{
\begin{array}{10}
f(x)f(y) & \mbox{if } f(x + y) \leq f(x)f(y)\\
f(x + y) & \mbox{otherwise}
\end{array}
\right.$$
National Math Camp 2020 Exam 2 Problem 2
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Re: National Math Camp 2020 Exam 2 Problem 2
Given, $f:\mathbb{R}\rightarrow\mathbb{R}$ such that $f(xy)=max\left(f(x+y), f(x)f(y)\right) \dots \dots \dots (1)$
Claim-1 : $0\leq f(0)\leq 1$
Proof :
Let's substitute $(x,y)=(0,0)$ in equation $(1)$,
We get, $f(0)=max\left(f(0),f(0)^2\right)$
$\rightarrow f(0)\geq f(0)^2\geq 0$, since $f(0)^2$ is always nonnegative.
$\rightarrow 0\leq f(0)\leq 1$ (Proved)
Claim-2 : If $f(x)\geq 0$ then $f(x)=f(0)$
Proof:
Since $0\leq f(0)\leq 1$ and $0\leq f(x)$,
So, $0\leq f(x)f(0)\leq f(x)$
$\rightarrow max\left(f(x), f(x)f(0)\right)=f(x)$
Let's substitute $y=0$ in equation $(1)$, we get the following:
$f(0)=max\left(f(x), f(x)f(0)\right)$
$\rightarrow f(0)=f(x)$ (Proved)
Claim-3 : If there exists $x\in\mathbb{R}$ such that $f(x)<0$, then $f(0)=0$
Proof :
Let's assume $x\in\mathbb{R}$ such that $f(x)<0$
Since $0\leq f(0)\leq 1$,
$\therefore 0\geq f(x)f(0)\geq f(x)$
$\rightarrow max\left(f(x), f(x)f(0)\right)=f(x)f(0)$
$\rightarrow f(0)=f(x)f(0)$ [By substituting $y=0$ in equation $(1)$, we get $f(0)=max\left(f(x), f(x)f(0)\right)$]
Now, $f(x)\neq 1$ since $f(x)<0$. So, it must be the case that $f(0)=0$. (Proved)
Notice that Claim-2 and Claim-3 implies that if $\exists a\in\mathbb{R}$ such that $f(a)<0$, then $\forall x\in\mathbb{R}$, $f(x)\leq0$.
So, let's just assume there exists $a\in\mathbb{R}$ such that $f(a)<0$.
$\therefore \forall x\in\mathbb{R}, f(x)\leq 0$
By substituting $y=x$ at equation $(1)$, we get,
$f(x^2)=max\left(f(2x),f(x)^2\right)$
Since $f(x)^2$ must be nonnegative, $f(x)^2\geq 0\geq f(2x)$.
So, $f(x^2)=max\left(f(2x),f(x)^2\right)=f(x)^2$.
Now, $f(x^2)\leq 0$ and $f(x)^2\geq 0$
So, $f(x^2)=f(x)^2=0$
$\rightarrow f(x)=0, \forall x\in\mathbb{R}$
But this contradicts the fact that $\exists a\in\mathbb{R}$ such that $f(a)<0$
$\therefore\forall x\in\mathbb{R}, f(x)\geq 0$
So, by Claim-2, $f(x)=f(0)$ for all real number $x$.
So, our solution is $f(x)=k$ for all $x\in\mathbb{R}$ where $0\leq k\leq 1$. And it is not hard to check that this conditon satisfies equation $(1)$
Claim-1 : $0\leq f(0)\leq 1$
Proof :
Let's substitute $(x,y)=(0,0)$ in equation $(1)$,
We get, $f(0)=max\left(f(0),f(0)^2\right)$
$\rightarrow f(0)\geq f(0)^2\geq 0$, since $f(0)^2$ is always nonnegative.
$\rightarrow 0\leq f(0)\leq 1$ (Proved)
Claim-2 : If $f(x)\geq 0$ then $f(x)=f(0)$
Proof:
Since $0\leq f(0)\leq 1$ and $0\leq f(x)$,
So, $0\leq f(x)f(0)\leq f(x)$
$\rightarrow max\left(f(x), f(x)f(0)\right)=f(x)$
Let's substitute $y=0$ in equation $(1)$, we get the following:
$f(0)=max\left(f(x), f(x)f(0)\right)$
$\rightarrow f(0)=f(x)$ (Proved)
Claim-3 : If there exists $x\in\mathbb{R}$ such that $f(x)<0$, then $f(0)=0$
Proof :
Let's assume $x\in\mathbb{R}$ such that $f(x)<0$
Since $0\leq f(0)\leq 1$,
$\therefore 0\geq f(x)f(0)\geq f(x)$
$\rightarrow max\left(f(x), f(x)f(0)\right)=f(x)f(0)$
$\rightarrow f(0)=f(x)f(0)$ [By substituting $y=0$ in equation $(1)$, we get $f(0)=max\left(f(x), f(x)f(0)\right)$]
Now, $f(x)\neq 1$ since $f(x)<0$. So, it must be the case that $f(0)=0$. (Proved)
Notice that Claim-2 and Claim-3 implies that if $\exists a\in\mathbb{R}$ such that $f(a)<0$, then $\forall x\in\mathbb{R}$, $f(x)\leq0$.
So, let's just assume there exists $a\in\mathbb{R}$ such that $f(a)<0$.
$\therefore \forall x\in\mathbb{R}, f(x)\leq 0$
By substituting $y=x$ at equation $(1)$, we get,
$f(x^2)=max\left(f(2x),f(x)^2\right)$
Since $f(x)^2$ must be nonnegative, $f(x)^2\geq 0\geq f(2x)$.
So, $f(x^2)=max\left(f(2x),f(x)^2\right)=f(x)^2$.
Now, $f(x^2)\leq 0$ and $f(x)^2\geq 0$
So, $f(x^2)=f(x)^2=0$
$\rightarrow f(x)=0, \forall x\in\mathbb{R}$
But this contradicts the fact that $\exists a\in\mathbb{R}$ such that $f(a)<0$
$\therefore\forall x\in\mathbb{R}, f(x)\geq 0$
So, by Claim-2, $f(x)=f(0)$ for all real number $x$.
So, our solution is $f(x)=k$ for all $x\in\mathbb{R}$ where $0\leq k\leq 1$. And it is not hard to check that this conditon satisfies equation $(1)$
"If people do not believe that mathematics is simple, it is only because they do not realize how complicated life is."
— John von Neumann
— John von Neumann
- FuadAlAlam
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Re: National Math Camp 2020 Exam 2 Problem 2
The answer is $f \equiv c$ where $0 \leq c \leq 1$. It is easy to check that these functions satisfy the given condiion. We will now show that they are the only solutions.
We divide it into $2$ cases.
Case 1 : $f(0) \leq f(0)^2$
Plugging $x = y = 0$, we have $$f(0) = f(0)^2$$
Thus, $f(0) = 0$ or $1$.
If $f(0) = 0$, we claim that $f(a) \leq 0$ for all real numbers $a$. Otherwise, there exists some real number $a$ such that $f(a) > 0$. Plugging $x = a, y = 0$, we get $f(a) = 0$, a contradiction.
Now, plugging $x = 1, y = -1$, we have $f(-1) = f(1)f(-1)$ or $f(-1) = 0$. As $f(1) \leq 0$, we have $f(-1) = 0$ in both cases.
Again, plugging $y = -1$, we get $f(-x) = 0$ as $f(x - 1) \leq 0$ for all real numbers $x$. Therefore, the only solution for $f(0) = 0$ is $f(t) = 0$ for all real numbers $t$.
If $f(0) = 1$, then plugging $y = 0$ yields $f(x) = 1$ for all real numbers $x$.
Case 2 : $f(0) > f(0)^2$
Obviously $f(0) > 0$. Thus, we have $0 < f(0) < 1$.
If $f(x) \leq f(x)f(0)$ for some real number $x$, plugging $y = 0$ yields $f(0) = f(x)f(0)$ or $f(x) = 1$. But this would imply $f(0) \geq 1$ from the fact that $f(x) \leq f(x)f(0)$, a contradiction.
Therefore, $f(x) > f(x)f(0)$ for all real numbers $x$.
Again, plugging $y = 0$ gives $f(x) = f(0)$. This solution satisfies the condition as $c > c^2$ for all $0 < c < 1$. Thus, $f \equiv c$ for $0 < c < 1$ in this case.
Therefore, the only solutions are constant functions $f \equiv c$ for $0 \leq c \leq 1$ and we are done.
We divide it into $2$ cases.
Case 1 : $f(0) \leq f(0)^2$
Plugging $x = y = 0$, we have $$f(0) = f(0)^2$$
Thus, $f(0) = 0$ or $1$.
If $f(0) = 0$, we claim that $f(a) \leq 0$ for all real numbers $a$. Otherwise, there exists some real number $a$ such that $f(a) > 0$. Plugging $x = a, y = 0$, we get $f(a) = 0$, a contradiction.
Now, plugging $x = 1, y = -1$, we have $f(-1) = f(1)f(-1)$ or $f(-1) = 0$. As $f(1) \leq 0$, we have $f(-1) = 0$ in both cases.
Again, plugging $y = -1$, we get $f(-x) = 0$ as $f(x - 1) \leq 0$ for all real numbers $x$. Therefore, the only solution for $f(0) = 0$ is $f(t) = 0$ for all real numbers $t$.
If $f(0) = 1$, then plugging $y = 0$ yields $f(x) = 1$ for all real numbers $x$.
Case 2 : $f(0) > f(0)^2$
Obviously $f(0) > 0$. Thus, we have $0 < f(0) < 1$.
If $f(x) \leq f(x)f(0)$ for some real number $x$, plugging $y = 0$ yields $f(0) = f(x)f(0)$ or $f(x) = 1$. But this would imply $f(0) \geq 1$ from the fact that $f(x) \leq f(x)f(0)$, a contradiction.
Therefore, $f(x) > f(x)f(0)$ for all real numbers $x$.
Again, plugging $y = 0$ gives $f(x) = f(0)$. This solution satisfies the condition as $c > c^2$ for all $0 < c < 1$. Thus, $f \equiv c$ for $0 < c < 1$ in this case.
Therefore, the only solutions are constant functions $f \equiv c$ for $0 \leq c \leq 1$ and we are done.