3rd Putnam 1940

Problem A3

α is a fixed real number. Find all functions f: R → R (where R is the reals) which are continuous, have a continuous derivative, and satisfy   ∫by fα(x) dx = ( ∫by f(x) dx )α for all y and some b.



Straightforward to get the basic idea, but care is needed with the details. It is quite hard to get the answer exactly right (the official solution gives a spurious solution for the case (2) - overlooking the problem with the integration limits).

(1) No solutions for α = 0;
(2) No solutions for α < 0;
(3) Any continuous f with continuous derivative is a solution for α = 1;
(4) For α > 0 and not of the form p/q, with p and q odd positive integers, f = A ekx, where A is any real, and k is the positive real value of α1/(α -1);
(5) For α = p/q, with p and q odd positive integers (but not both 1), f = A ekx or A e-kx, where A is any real, and k is the positive real value of α1/(α -1).

Case (3) is obvious. Case (1) is almost obvious: if α = 0, then the lhs varies with y ( = y - b), but the rhs does not ( = 1), so there are no solutions. Assume now that α is not 0 or 1.

Differentiate wrt y and put g(y) = ∫by f(x) dx. Then f = g' and we have fα = α gα -1 f. So f = k g, where k = α1/(α -1). Since f = g', we can integrate immediately to get f(x) = A ekx (*).

However, we have to consider how many real values k can have. If α > 0, then k certainly has a positive value, but we can also take the corresponding negative value if 1/(α - 1) involves an even root, in other words if α = p/q with p and q both odd. Finally, (*) is clearly necessary, but not necessarily sufficient, so we have to subsitute (*) back into the original equation. For k > 0, we find the solution works provided b = -∞. If k < 0, then it works with b = ∞. This gives cases (4) and (5). [Note, however, that we can only allow A negative for α not of the form p/q with p and q odd, provided we are content for both sides of the original equation to have complex values (even though f is real valued).]

For α < 0, all values of α1/(α -1) are complex unless α = - p/q with p an even positive integer and q an odd positive integer. If α has that form, then we may take k = - 1/|α|1/(|α|+1). But now there is a problem with b. Taking A = 1 for simplicity, the fα(x) = e|αk|x, so the lhs = const (e|αk|y - e|αk|b) and the rhs = const /(e- |k|y - e- |k|b)|α|. The constants are the same, but we need b = -∞ to get rid of the b term on the lhs and b = ∞ to get rid of the term on the rhs. So there are no solutions in case (2).  


3rd Putnam 1940

© John Scholes
14 Sep 1999