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I want to know how scientists know that the inner product of f and g equal to integration from $$\int_a^b f(x)g(x)\ dx.$$

amWhy
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nari
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1 Answers1

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An inner product on a vector space $V$ over $\mathbb R$ is a function $\langle\cdot,\cdot\rangle : V \times V \to \mathbb R$ which satisfies certain axioms, e.g., $\langle v, v\rangle = 0$ iff $v = 0$, $\langle v, v\rangle \geq 0$ for all $v$.

The definition of an inner product on the vector space of let's say continuous functions on $[0,1]$ of

$$\langle f, g \rangle = \int_0^1 f(x)g(x) \ dx$$

works in that it satisfies those axioms.

Now, given that, it turns out we can use the inner product for a very wide range of useful things.

Historically, the definition was probably inspired by analogy with vectors in $\mathbb R^n$, namely

$$\langle (v_1,v_2,...,v_n),(w_1,w_2,...,w_n)\rangle = \sum_{i=1}^n v_iw_i$$

Simon S
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  • yes , but how this sigma turns to integration form this is my question. – nari Feb 11 '15 at 18:16
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    By analogy with a Riemann sum. Suppose we sample the functions $f$ and $g$ at $1/n, 2/n, ..., (n-1)/n, 1$. Then a draft of our inner product could be

    $$R_n(f,g) = \sum_{i=1}^n f(i/n)g(i/n)$$

    We had better normalize this for $n$ otherwise it will run away from us

    $$S_n(f,g) = \frac{1}{n} \sum_{i=1}^n f(i/n)g(i/n)$$

    Now let $n\to\infty$ so we get closer to capturing all of the information in $f$ and $g$. Then $S_n \to \int_0^1 fg$

     – Simon S Feb 11 '15 at 18:20
  • sorry, I don't understand you, could you please give me another simple explanation or the complete origion prove , I'am just a student, thank you. – nari Feb 11 '15 at 18:29
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    Which part don't you understand: the definition of $R_n$, or $S_n$, or the limit $S_n \to \int fg$? – Simon S Feb 11 '15 at 18:39
  • I don't understand the Riemann sum. I don't listen about it before. – nari Feb 11 '15 at 18:41
  • Riemann sums is how we define the Riemann integral, the integrals you have used in your life up to now: https://www.khanacademy.org/math/integral-calculus/indefinite-definite-integrals/definite_integrals/v/riemann-sums-and-integrals – Simon S Feb 11 '15 at 18:47
  • ok I understand the video but please could you tell me what's meant by that: By analogy with a Riemann sum. Suppose we sample the functions f and g at 1/n,2/n,...,(n−1)/n,1. Then a draft of our inner product could be Rn(f,g)=∑i=1nf(i/n)g(i/n) We had better normalize this for n otherwise it will run away from us Sn(f,g)=1n∑i=1nf(i/n)g(i/n) Now let n→∞ so we get closer to capturing all of the information in f and g. Then Sn→∫10fg – nari Feb 11 '15 at 20:51
  • The expression $S_n$ is a Riemann sum. The idea is that as $n$ gets larger and larger, the sum captures more values of the functions $f$ and $g$. In the limit, $S_n$ converges to the integral. The integral then has all the nice inner product properties. The sums $R_n$ and $S_n$ do not. – Simon S Feb 11 '15 at 21:03
  • ok but , how n tends to infinite and from the relation of Sn=1/n*gima , so when n tends to infinite , Sn tends to zero ? – nari Feb 11 '15 at 21:21
  • No. $S_n$ is a Riemann sum for the integral $\int_0^1 f(x)g(x) \ dx$. In the limit as $n\to\infty$, $S_n$ converges to that integral. – Simon S Feb 11 '15 at 21:41
  • ok,thank you simon for helping me. – nari Feb 11 '15 at 21:50
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    @SimonS why do you need to normalise $R_n$ to $S_n$? And how you do it? – newbie125 Nov 04 '16 at 14:34
  • What if the field is $\mathbb{C}$ instead? Then we won't have the integral satisfy the definition of inner product, correct? – information_interchange Mar 12 '20 at 20:52
  • @information_interchange you can loosen the symmetric property and redefine it with antisymmetric property – user29418 Feb 15 '23 at 17:26