Re: Integration as functional

AARGH!

Stephen J. Herschkorn wrote:

Let X be an arbitrary set; let F denote the real or complex field, and let D be a subalgebra of F^X


Another correction! D should be a linear subspace, not an algebra.

with the properties
(i) if f in D, then |f| in D, and
(ii) if f in D, g in F^X, and |g| <= f, then g in D.

Let I in F^D be linear and positive (i.e., if f is nonnegative, then so is If).

Let p be greater than 1 and f in D be nonnegative. If f^p in D and I(f^p) = 0, is it necessarily true that
f in D and If = 0?


The background:

I got it into my head to see what exactly is needed for thew proof of the Minkowski inequality. I think it is bolied down to the first two paragraphs above. More explicitly, for p >= 1, let L^p = {f in F^X: |f|^p in D}. Then the above set up implies that L^p is an algebra. For f in L^p, define ||f||_p = [I(|f|^p)]^(1/p). Then the above is enough to show that if
f, g in L^p, then ||f + g||_p <= |||f||_p + ||g||_p. (Actually, looking at the details, I don't think I even use (i).) To me, stripping it down like this makes the proof much clearer (even for establishing the p-norm on R^n).

This implies that ||.||_p is a seminorm (or pseudonorm, if you prefer). We can define the equivalence relationship ~p on L^p by f ~p g iff ||f - g||_p = 0. Then the collection of equivalence classes with ||.||_p is a well-defined noremd vector space. My question above is asking if one can conclude that ~p is the same as ~1 for any p > 1, as is the case with usual abstract integration.

You can see that the above properties get you fairly far. Without (i), one can show:
- f, g in D, f <= g implies If <= Ig;
- |f| in D implies f in D, and
- |f| in D implies |If| <= I|f|.
Throw in (i), you also get
- if f in D and f is real-valued, then If is real;
- if f in D, then so are Re f and Im f, and Re If = I(Re f), Im If = I(Im f), and
- I[conj(f)] = conj(If), where conj denotes the complex conjugate.



--
Stephen J. Herschkorn sjherschko@xxxxxxxxxxxx
.

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