求满足函数方程$f(xf(y))=yf(x)$的所有$f:\mathbb{R}^+\rightarrow\mathbb{R}^+$函数

Let's start by defining a shorthand for the given functional equation: let ( P(x,y) ) represent the statement ( f(xf(y)) = yf(x) ) for positive real numbers ( x,y ).

Step 1: Prove ( f(1) = 1 ) #

First, set ( x = 1 ) in ( P(x,y) ):

( f(f(y)) = yf(1) ) --- (1)

Next, set ( y = 1 ) in the original equation (( P(x,1) )):

( f(xf(1)) = f(x) ) --- (2)

From equation (1), we can see ( f ) is surjective: for any ( z \in \mathbb{R}^+ ), ( f(f(z/f(1))) = z ), meaning every positive real is in the image of ( f ). Since ( f ) is surjective, applying ( f ) to both sides of equation (2) gives ( xf(1) = x ) for all ( x \in \mathbb{R}^+ ), so ( f(1) = 1 ).

Step 2: Show ( f ) is an involution (( f(f(x)) = x )) #

Substitute ( f(1) = 1 ) back into equation (1):

( f(f(y)) = y ) for all ( y \in \mathbb{R}^+ )

This tells us ( f ) is its own inverse, so it's bijective (both injective and surjective).

Step 3: Prove ( f ) is multiplicative (( f(xy) = f(x)f(y) )) #

Now consider ( P(x, f(y)) )—that is, replace ( y ) with ( f(y) ) in the original functional equation:

( f(xf(f(y))) = f(y)f(x) )

But since ( f(f(y)) = y ), this simplifies directly to:

( f(xy) = f(x)f(y) ) for all ( x,y \in \mathbb{R}^+ )

Step 4: Find all solutions #

We now have a bijective multiplicative function mapping positive reals to positive reals. For any rational number ( q ), we can extend the multiplicative property to show ( f(x^q) = f(x)^q ) (for example, ( f(x^n) = f(x)^n ) for integers ( n ), and ( f(x^{1/n}) = f(x)^{1/n} ) since ( f ) is bijective).

The standard bijective multiplicative functions on ( \mathbb{R}^+ ) are of the form ( f(x) = x^k ) where ( k ) is a real constant. Let's substitute this into the original equation to find valid ( k ):

• Left-hand side: ( f(xf(y)) = (x \cdot y^k)k = x^k y^{k2} )
• Right-hand side: ( yf(x) = y \cdot x^k )

For these to be equal for all ( x,y \in \mathbb{R}^+ ), we need ( k^2 = 1 ). This gives two solutions: ( k = 1 ) or ( k = -1 ).

Checking these:

• If ( k=1 ), ( f(x) = x ): ( f(xf(y)) = f(xy) = xy = yf(x) ), which holds.
• If ( k=-1 ), ( f(x) = 1/x ): ( f(xf(y)) = f(x \cdot 1/y) = y/x = y \cdot 1/x = yf(x) ), which also holds.

Conclusion #

All solutions to the functional equation are:

• ( f(x) = x )
• ( f(x) = \frac{1}{x} )

内容的提问来源于stack exchange，提问作者Bless