A function h(x) is positive or zero for all values of x. Assume h(x)
is even. If the Fourier transform of h(x) is H (u) show that,

For all u, IH (u) 1 <= IH (O) I.

Show that as u ~ 00 IH (u) 1 ~ O.

Please provide the detail steps involved as I have to understand the problem. Thanks

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Clarification of Question by aathi-ga on 14 Dec 2004 22:09 PST

Sorry for this mis understanding. Due the reason that google wont
allow to attach a word file so its some time become difficult to write
the exact expressions. Anyway here is the question once again and I
hope it will be clear this time.

A function h (x) is positive or zero for all values of x. Assume h(x)
is even .If the Fourier transformation of h(x) is H(u) show that,

For all u, |H(u)| <= |H(0)|

show that as  u -> infinity as |H(0)| -> 0

Please note " <= " means this implies
            " -> " means approaches to
            "|H(0)|" means Mode of H(0) 

I hope things should be clear now.

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Request for Question Clarification by mathtalk-ga on 16 Dec 2004 17:07 PST

Hi, aathi-ga:

Your restatement was quite helpful, but if I may, I'll put it into a
slightly different form:

Let h(x) be a nonnegative real function and even, and let H(u) be the
continuous Fourier transform of h(x).  Show that:

(1) |H(u)| <= |H(0)|  for all u

(2) 0 = LIMIT |H(u)| as u --> +oo

We remark that |H(u)|, the absolute value of H(u), is also called the
"modulus" of H(u).  Thus (2) is equivalent to H(u) itself tending to
zero as u approaches plus infinity; if the absolute value of a complex
number is near zero, then the complex number is near zero.

regards, mathtalk-ga

Your notation makes understanding the question slightly difficult.  I
think for the first part you are trying to show, where I is an
integral, that:

IH(u)du <= IH(0)du for all u.

h(x) is even, which means that h(x) = h(-x) for all x.  I think this
carries through to the function H, so that H(u) = H(-u).  Which means
that you only need to solve the problem for u =>0.

H(u) = Ih(x)e^(2PI)iux dx

In this equation u=0 implies e^(2PI)iux = 1

So H(0) = Ih(x) dx >= Ih(x)e^(2PI)iux dx since the modulus of e^(2PI)iux is 1.

So H(0) >= H(u), so IH(0) >= IH(u)

I may have got it wrong, or possibly misunderstood the question.