Inner Product

2008-03-18 5:32 am
Let V = { v∈V : v∈(C^0[a,b] )∩(C^2]a,b[ ), v(a)=v(b)=0 }, it is not difficult to show that V is a real vector space.

i) Show that ∫(-u''v + cuv) dx = ∫(u'v' + cuv) dx for all u, v ∈V
where the integration is taken from a to b and c is a positive constant.

ii) Show that <u, v> = ∫(u'v' + cuv) dx is an inner product for V.

The above has just shown that V is an inner product space equipped with the inner product <‧,‧>
更新1:

好快,我差點以為我在自問自答。 In fact, I feel something wrong with the above questions. I consider I may have missed some important detail(s) in it after thinking twice. Please let me know if there is any assumption(s) I have to make in order to solve the above problem.

更新2:

For Axioms 1 , it would be better to clarify that the integral involved are all bounded and hence well-defined since they are continuous on the closed interval [a , b] , otherwise we can't split the integral into two. Axioms 4 needs to be clarified more.......

更新3:

BTW, if everything is finally correct, could we reduce some condition in the question?

回答 (3)

2008-03-18 7:20 am
✔ 最佳答案
上面的同學冇答問題喎:


圖片參考:http://i187.photobucket.com/albums/x22/cshung/7008031703220.jpg


References come from Mathworld - Inner Product

2008-03-18 23:23:55 補充:
For the proof of the last axiom

http://s187.photobucket.com/albums/x22/cshung/?action=view&current=70080317032201.jpg
參考: 從不抄襲。, 從不抄襲。
2008-03-19 6:44 pm
Axiom 1: continuous + compact interval => bounded. So boundedness is implied by the question.
Axiom 4 is totally fine in the answer... no need clarified more ... Or considering Lebesgue integration, can conclue v = 0 a.e., and v is continuous => v = 0 everywhere.

2008-03-19 10:48:53 補充:
For the conditions, they are all used in the answer to i) only.
C^2 is needed for u'' to make sense
v(a) = v(b)=0 is used for the integration by parts. Here we can weaken a bit and say v(a)=v(b) is enough. This still define a vector space.

2008-03-19 10:48:58 補充:
For ii), it works for all function in C^0[a,b] ∩ C^1]a,b[ --- we just need the continuity of u and u' only. Refer to Andrew's proof.
2008-03-18 5:36 am
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Inner Product



An inner product is a generalization of the dot product. In a vector space, it is a way to multiply vectors together, with the result of this multiplication being a scalar.

More precisely, for a real vector space, an inner product satisfies the following four properties. Let , , and be vectors and be a scalar, then:

1. .

2. .

3. .

4. and equal if and only if .

A vector space together with an inner product on it is called an inner product space. This definition also applies to an abstract vector space over any field.

Examples of inner product spaces include:

1. The real numbers , where the inner product is given by

(1)

2. The Euclidean space , where the inner product is given by the dot product

(2)

3. The vector space of real functions whose domain is an closed interval with inner product

(3)

When given a complex vector space, the third property above is usually replaced by

(4)

where refers to complex conjugation. With this property, the inner product is called a Hermitian inner product and a complex vector space with a Hermitian inner product is called a Hermitian inner product space.

Every inner product space is a metric space. The metric is given by

(5)

If this process results in a complete metric space, it is called a Hilbert space.

SEE ALSO: Complete Metric Space, Dot Product, Hermitian Inner Product, Hilbert Space, Inner Product Space, Interior Product, L2-Inner Product

This entry contributed by John Renze




CITE THIS AS:

Renze, John. &quot;Inner Product.&quot; From MathWorld--A Wolfram Web Resource, created by Eric W. Weisstein. http://mathworld.wolfram.com/InnerProduct.html

2008-03-18 13:52:51 補充:
sorry,有D電惱答5到
參考: myself


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