Problem of the Day: June 14

Let f be a function on the reals so that for every x,y we have f(xf(y))=yf(x). Prove that f is odd.

Edited to add solution: (This solution combines ideas from Nir's and Saskia's solutions -- it starts using Nir's proof of involutivity, then uses Saskia's proof of multiplicative distribution, and then follows both of them in calculating f(-1).)

Step 1: f is involutive. We have f(xf(y))=yf(x). Applying f to both sides, we have f(f(xf(y)) = f(yf(x)). But f(yf(x))=xf(y), from our original condition swapping x and y. Thus f(f(xf(y))=xf(y), so f is involutive (that is, f is its own inverse).

Involutive is a powerful condition worth looking out for in functional equation problems. Involutive functions are always bijective, and this fact is frequently what we use to go forward.

Step 2: f distributives over multiplication. Substituting f(x) for x we have f(f(x)f(y)) = yf(f(x) = xy (by involutivity). Now take f of both sides, and we have f(f(f(x)f(y)) = f(xy). Applying involutivity to the left-hand side we have f(x)f(y)=f(xy).

Step 3: We calculate f(1). Setting x=1 we have f(f(y)) = yf(1). By involutivity, y = yf(1), so f(1) = 1.

Step 4: We calculate f(-1). Since f is involutive, there is k such that f(-1)=k and f(k)=-1. Then 1 = f(1) = f(-1*-1) = f(-1*f(k)) = (by our given condition) kf(-1) = k^2. So k^2 = 1, and k = 1 or k = -1. But since f(1) = 1, we cannot have f(-1) = 1 given that f is bijective. Thus k=-1, and f(-1) = -1.

Step 5: We show that f is odd. f(-a) = f(-1*a) = f(-1)*f(a) (by multiplicative distribution) = -f(a).

(Problem source: 2012 Kazakhstan National Olympiad Grade 11 Day 2 Problem 1)

Comments

  1. Nir ElberJune 15, 2019 at 12:34 PM

    I have a relatively uninteresting solution. Wander aimlessly until done.

    Let the assertion be P(x,y). If f = 0, we are done. Else, there is a nonzero element f(a).

    f is an involution (and thus also bijective):
    P(1,y) gives f(f(y)) = y f(1). Tripling the involution tells us that
    f(y f(1)) = f(f( f(y) )) = f(y) f(1).
    However, P(y,1) tells us that f(y f(1)) = f(y), so y=a gives us that f(a) = f(a)f(1) and f(1) = 1.
    Plugging this in to P(1,x) gives us that f(f(x)) = x. As desired.

    f is odd:
    Suppose f(-1) = k so that f(k) = -1 as well. Dropping P(-1, k) tells us that
    f(1) = f(-1 * -1) = f(-1 f(k)) = k f(-1) = k^2.
    Thus, k^2 = 1 and k = 1 or -1. Because f is bijective (involution), k = -1 as f(1) =/= f(-1). Therefore f(-1) = -1.
    Finally, dropping P(x,-1) tells us that
    f(-x) = f(x * -1) = f(x f(-1)) = - f(x).

    As desired.

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  2. Saskia SolotkoJune 15, 2019 at 12:55 PM

    First of all, by symmetry, we have f(yf(x))=xf(y), so
    f(f(xf(y))=f(yf(x))=xf(y). This is basically saying that if f(xf(y))=yf(x) is true for some ordered pair (x,y), it is also true for (y,x).

    Now we can divide the problem into two cases:

    Case 1: f(y)=0 for all x: f(y)=0 is the only solution so in this case f is odd. Not difficult.

    Case 2: f(y) is nonzero for some y: We have f(f(xf(y))=xf(y). Let f(y)= some nonzero constant c, and let h(x)=cx=a. Since h(x) is a bijection we have f(f(a))=a for all a, or f is an involution. (Basically, if f(y) is nonzero for some y, then we can get xf(y) to be any number)

    Since f is an involution, it is trivially a bijection.

    Now that we know f is a bijection, we can do cool stuff.

    Let x=f(t): Now we have that f(f(t)f(y))=yf(f(t)) (from the first equation). Simplifying, we get:

    f(f(x)f(y))=xy

    f(f(f(x)f(y)))=f(xy)

    Using the fact that f is involutive, we get:

    f(x)f(y)=f(xy). So f is multiplicative! Nice!

    Since f is involutive, we have f(f(1))=1, and from the first equation we have f(1f(1))=1f(1) or f(f(1))=f(1). So we know f(1)=1.

    Since f is multiplicative, we have f(-1)f(-1)=f(1)=1, or (f(-1)^2)=1. 1 has two square roots: -1, and 1. Since f is injective and f(1)=1, f(-1)=-1.

    Now we can finish:

    Once again using f(a)f(b)=f(ab), we have:

    f(b)f(-1)=f(-b), or -f(b)=f(-b). So f must be odd.

    Solution by Saskia Solotko

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