If f and g are positive continuous functions on [0,1], k is a positive continuous kernel on [0,1]x[0,1],...

analysis/integral.kernel New

New entry, edited from material collected for the archive and not yet in a released version.

Problem

If f and g are positive continuous functions on [0,1], k is a positive
continuous kernel on [0,1]x[0,1], and each of f and g is the integral
transform of the other by k, prove that f = g.

Notation:  I = [0,1]  the closed unit interval;
           R+ = the positive real numbers
           Int_{a}^{b} means the integral from a to b

The hypotheses:  f,g : I --> R+  are continuous functions
                   k : IxI --> R+  is a continuous function of two variables
                 Int_{0}^{1}f(y)k(x,y)dy = g(x)  for all x in I
                 Int_{0}^{1}g(y)k(x,y)dy = f(x)  for all x in I

Prove:  for all x in I,  f(x) = g(x).

-- Frank Pinto <rolls@cis.umassd.edu>, February 1994, who says it
   appeared in the Putnam competition

[ Editor: the first paragraph is the editor's summary.  The year of the
Putnam competition was not given. ]

Show solution

[ Editor: an earlier attempt by David Ring and David Savitt's objection
to it were not in the collected material. ]

David Savitt <dsavitt@unixg.ubc.ca> wrote:
>So, your argument induces no contradiction.

You're right, I lost a sign there. But remarkably, I think I can patch
it up.

Assume f(x) != g(x)*a .

Let r be max[f(x)/g(x)] (occuring at x=x0). If r <= 1 switch f and g.

Now consider Int_{0}^{1}[r*g(y)-f(y)]k(x,y)dy = r*f(x)-g(x) .

rg-f >= 0 and has at least one non-zero point. So rf-g > 0 everywhere.
It is continuous and its domain is closed, so it has a minimum.

Then there must exist eps > 0 such that (r-eps)f-g > 0 everywhere.

Consider Int_{0}^{1}[(r-eps)*f(y)-g(y)]k(x0,y)dy = (r-eps)*g(x0)-f(x0) < 0 .

But the integrand is positive, contradiction.

I hope this one is correct :-)

-- Dave Ring <dwr2560@TAMUTS.TAMU.EDU>

Better yet:

Use operator notation: Kf = Int_{0}^{1} f(y)k(x,y)dy

>Assume f(x) != g(x)*a .
>Let r be max[f(x)/g(x)] (occuring at x=x0). If r <= 1 switch f and g.

KK(rg-f) = rg-f .

On the left side, rg-f >= 0 and has at least one non-zero point so the
right side is positive everywhere. But it has at least one zero.
Contradiction.

-- Dave Ring <dwr2560@TAMUTS.TAMU.EDU>