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Algebra Difficulty 7.3 National olympiad, round 2 Find the answer

Let R>0\mathbb{R}_{>0} be the set of all positive real numbers. Find all strictly monotone (increasing or decreasing) functions f:R>0Rf:\mathbb{R}_{>0} \to \mathbb{R} such that there exists a two-variable polynomial P(x,y)P(x, y) with real coefficients satisfying
f(xy)=P(f(x),f(y)) f(xy)=P(f(x), f(y))
for all x,yR>0x, y\in\mathbb{R}_{>0}.\\

A number or a short expression. Spacing and $ signs are ignored.

Solution

To solve the problem of finding strictly monotone functions f:R>0R f: \mathbb{R}_{>0} \to \mathbb{R} that satisfy the given equation f(xy)=P(f(x),f(y)) f(xy) = P(f(x), f(y)) for some two-variable polynomial P(x,y) P(x, y) , we'll proceed as follows:

Step 1: Analyze the Problem

We are given that f f is a strictly monotone function and P(x,y) P(x, y) is a polynomial such that:
f(xy)=P(f(x),f(y)) f(xy) = P(f(x), f(y))
for all x,yR>0 x, y \in \mathbb{R}_{>0} .

**Step 2: Possibilities for f f **

Since f f is strictly monotone, it can either be strictly increasing or strictly decreasing. Several forms of strictly monotone functions might be considered, and we need to check for consistency with the given functional equation.

1. Logarithmic Form: Suppose f(x)=clog(x) f(x) = c \log(x) , where c0 c \neq 0 . Then:
f(xy)=clog(xy)=c(log(x)+log(y))=clog(x)+clog(y)=P(clog(x),clog(y)) f(xy) = c \log(xy) = c (\log(x) + \log(y)) = c \log(x) + c \log(y) = P(c \log(x), c \log(y))
If we let P(u,v)=u+v P(u, v) = u + v , this form is compatible with the condition.

2. Power Form: Suppose f(x)=cxk f(x) = cx^k , where c0 c \neq 0 and k0 k \neq 0 . Then:
f(xy)=c(xy)k=cxkyk=P(cxk,cyk) f(xy) = c(xy)^k = cx^k y^k = P(cx^k, cy^k)
If we let P(u,v)=uv P(u, v) = uv , this form is also compatible with the condition.

Step 3: Verify the Monotonicity

The forms f(x)=clog(x) f(x) = c \log(x) and f(x)=cxk f(x) = cx^k are strictly monotone if the constants c c and k k are chosen appropriately:

- For f(x)=clog(x) f(x) = c \log(x) : The function is strictly increasing if c>0 c > 0 and strictly decreasing if c<0 c < 0 .

- For f(x)=cxk f(x) = cx^k : The function is strictly increasing if c>0 c > 0 and k>0 k > 0 , or strictly decreasing if c<0 c < 0 and k>0 k > 0 .

Conclusion

The strictly monotone functions f(x) f(x) that satisfy the functional equation are:
f(x)=clog(x)orf(x)=cxk f(x) = c \log(x) \quad \text{or} \quad f(x) = cx^k
where c c and k k are constants with suitable restrictions to maintain monotonicity. Thus, the solution is:
f(x)=clog(x) or f(x)=cxk \boxed{f(x) = c \log(x) \text{ or } f(x) = cx^k}

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Source: Omni-MATH, licensed Apache-2.0. Statement and solution reproduced as published; topic and difficulty added by this site.