### 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.

Solution

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).