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Vanishing Theorem
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3X.lich
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#1
Vanishing Theorem
Vanishing Theorem: Let f(x) be a continuous function in a finite closed interval [a,b]. Assume that f(x) \geq 0 in the interval and that \int_a^b f(x) dx=0. Then f(x) vanishes, i.e. f is identically zero.

Proof: The key point to the theorem is that a positive function above the x-axis would have zero area under the curve, but is that possible?. So suppose that f(c)>0 for some c \in (a,b). Let \varepsilon = f(c)/2 since f is continuous at c. Then there exists a \delta >0 such that |x-c|< \delta implies that f(x)> \varepsilon. Now the area under the curve over the interval (a,b) covers a rectangle with height \varepsilon and width 2\cdot \delta so in terms of integrals we have

\int_a^b f(x) dx \geq \int_{c- \delta}^{c+ \delta} f(x) dx \geq \varepsilon \cdot (2\cdot \delta )= \delta f(c)>0. We have a contradiction on the assumption that the integral is zero!!! Therefore f(x)=0 for all a<x<b. Since f is continuous, f(a)=f(b)=0 as well!!!
End of proof.

This theorem can be extended to higher dimensions (several variables) and perhaps the following theorem is beautiful as well:

Theorem: Let f(\mathbf{x}) be a continuous mapping in \mathbf{D}_0 such that \int \int \int_{\mathbf{D}} f(\mathbf{x}) d\mathbf{x} = 0 for all subdomains \mathbf{D} a subset of D_0. Then f(\mathbf{x})=0 in \mathbf{D}_0.
Hint for proof: Let \mathbf{D} be an open ball and have its radius r\rightarrow 0. Smile
Edit 11/21: Corrected the \LaTeX formulas
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Last edited by 3X.lich on Sat Nov 20, 2004 11:58 pm; edited 2 times in total 
PostPosted: Fri Mar 05, 2004 9:50 pm  Back to top 
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sjk
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#2
I think there is a more simple proof.

Take F(x)=\int_a^xf(x)\;dx. Then 0\leq F(x)\leq\int_a^bf(x)\;dx=0, so F is identically zero. Finally, 0=F'(x)=f(x).

PostPosted: Sat Nov 20, 2004 8:44 am  Back to top 
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Myth
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#3
Re: Vanishing Theorem
3X.lich wrote:
Theorem: Let f(x) be a continuous mapping in D_0 such that \int_D f(x)dx=0 for all subdomains D a subset of D_0. Then f(x)=0 in D_0.

We don't need continuity in this theorem, but only f\in L_{1,loc}(D_0).
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PostPosted: Sat Nov 20, 2004 9:40 am  Back to top 
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Kent Merryfield
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Myth wrote:
We don't need continuity in this theorem, but only f\in L_{1,loc}(D_0).

That's a different theorem. If you drop the condition of contiunity, then you cannot conclude that f is zero everywhere, as would be shown by the trivial example of a function that is 1 at one point and zero everywhere else. The theorem Myth is referring to has as its conclusion that f is zero almost everywhere, or, equivalently, that it is the zero element of L^1. But the elements of L^1, although often informally referred to as functions, are not functions; they are equivalence classes of functions under the equivalence relation of almost everywhere equality.

For continuous functions, the proof that 3X.lich posted is basically what has to be done, except that he made the too-strong assumption that f\ge 0. Although you need something like that in the proof, you don't have to assume it up front.

Here's another way to put it, letting f be complex-valued. Suppose f is a continuous complex-valued function on D_0 such that for any (open) domain D\subset D_0, \int_Df(x)dx=0. Suppose that for some x_0, f(x_0)\ne0. Let \alpha=\frac{\overline{f(x_0)}}{|f(x_0)|}. Then \alpha f(x_0)=|f(x_0)|. By the continuity of f (and hence Re(\alpha f)), then there is some neigborhood B of x_0 on which Re(\alpha f(x))>\frac{|f(x_0)|}2. Hence Re\left(\int_B\alpha f(x)dx\right)\ge|B|\frac{|f(x_0)|}2>0 and thus \int_B f(x)dx\ne0. Since this cannot happen, we conclude that f can never take on a nonzero value.

PostPosted: Sat Nov 20, 2004 6:29 pm  Back to top 
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Myth
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#5
Kent Merryfield wrote:
Myth wrote:
We don't need continuity in this theorem, but only f\in L_{1,loc}(D_0).

That's a different theorem. If you drop the condition of contiunity, then you cannot conclude that f is zero everywhere, as would be shown by the trivial example of a function that is 1 at one point and zero everywhere else. The theorem Myth is referring to has as its conclusion that f is zero almost everywhere, or, equivalently, that it is the zero element of L^1. But the elements of L^1, although often informally referred to as functions, are not functions; they are equivalence classes of functions under the equivalence relation of almost everywhere equality.

It is clear that if I say f\in L_{1,loc} then identity f=0 has meaning in L_1-sense. But if you take continuous representative of class 0\in L_{1,loc} it must be 0 everywhere.
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PostPosted: Sat Nov 20, 2004 10:44 pm  Back to top 
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3X.lich
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#6
Re: Vanishing Theorem
Myth wrote:
3X.lich wrote:
Theorem: Let f(x) be a continuous mapping in D_0 such that \int_D f(x)dx=0 for all subdomains D a subset of D_0. Then f(x)=0 in D_0.

We don't need continuity in this theorem, but only f\in L_{1,loc}(D_0).

It is possible for the theorem to hold without the continuity assumption in more generalized spaces. I'll check back later.

Kent Merryfield: you are right about the assumption that f\geq 0 is too strong. I wanted to present a proof accessible to beginning calculus students that may find several variables hard.
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PostPosted: Sat Nov 20, 2004 11:54 pm  Back to top 
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