3.3 THEORY OF FUNCTIONS
195
Remark. One can prove with the same method the generalization that one obtains by allowing different functions fk on the left of the assumed equality.
Solution 2. In the second solution, we will not use the existence of the mean m; otherwise, the argument that follows is similar to the one above. We start from the identity n
E Ck
f(Xyk)X(Xyk)Xk(y) = f(x)X(x)g(y)
k=1
that we obtained in the first solution in (1). Applying this with x = y-11 ... , x = y-m and adding the resulting equalities together, we obtain with
m
Sm(y) = 1: f (xy-i)X(xy i) j=1
the identity
E ck
m
k-1
n
(Sm(Y)
k=1
+ j=0 E -j=m-k+1 E
f (xya)X(xyi)
k(y) = Sm(y)g(y) (3)
By the assumption, Re Sm (y) > me -> oo
as m -> oo,
where c denotes a positive lower bound on the real parts of the products f (x)X(x). Hence, the numbers ISm(y)I tend to infinity as m -* oo. Now if we divide (3) by Sm(y) and let l tend to infinity, then we again arrive at (2) in view of the boundedness of f. This completes the proof.
Problem F.27. Suppose that the components of the vector . . . , u,,) are real functions defined on the closed interval [a, b] with the property that every nontrivial linear combination of them has at most n zeros in [a, b]. Prove that if a is an increasing function on [a, b] and the rank of the operator
u=(u0,
b
A(f) =
Ja
u(x) f (x)do(x),
f E C[a, b],
is r < n, then a has exactly r points of increase. Solution. By the assumption that there is a 0 i4 c E R'°+1 orthogonal onto the range of A, that is, (c, A(f )) = 0 for every f E C[a, b]. This means that
Ja
b(c, u(t))
f (t)da(t) = 0