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I've searched for the inner product definition and I saw that it should only satisfy some conditions (axioms) and thus there could be several operations representing an inner product, one of which is the usual multiplication, or the dot product. the question is: why is the inner product of 2 functions is defined by $\int f_1(x)f_2(x)dx$? their choice of the operation is based on what? there could be other operations satisfying the inner product conditions so why this ?

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  • $\begingroup$ On a related question how do you do the inner product < > in LaTeX? $\endgroup$ – Alec Teal Feb 13 '15 at 0:56
  • $\begingroup$ \langle blah blah \rangle $\endgroup$ – Ittay Weiss Feb 13 '15 at 0:56
  • $\begingroup$ @IttayWeiss cheers. $\endgroup$ – Alec Teal Feb 13 '15 at 0:58
  • $\begingroup$ @AlecTeal You should have asked your question on tex.stackexchange.com, or at the very least in a separate question. Asking questions within the comments of a different (and quite unrelated) question isn't good style - it makes your question and its answers very hard to find for others. $\endgroup$ – fgp Feb 13 '15 at 1:18
  • $\begingroup$ @fgp It was a tiny question and asking "how do I do [something related to post]" in comments here is how I learnt. $\endgroup$ – Alec Teal Feb 13 '15 at 1:31
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The history of inner product began with a problem of being given a function $f$ and trying to determine constants $a_{n}$, $b_{n}$ such that $$ f(x) = a_{0}+a_{1}\cos(x)+b_{1}\sin(x)+a_{2}\cos(2x)+a_{3}\sin(2x)+\cdots, \;\;\;\; 0 \le x \le 2\pi. $$ This came up in solving for vibrations of a string where the initial displacement of the string was $y=f(x)$. Amazingly, Euler and Clairaut discovered around 1775 that if you multiplied by one of the functions $\{ 1,\cos(x),\sin(x),\cos(2x),\sin(2x),\cdots\}$ and integrated over $[0,2\pi]$, all of the integrals on the right would be $0$ except for the one involving the function you multiplied by. This gave a way to isolate the coefficients $a_{n}$ and $b_{n}$. This trick was further exploited by Fourier in solving his Heat Equation using separation of variables. Fourier discovered orthogonality conditions for all kinds of equations.

Weighted integral expressions naturally arose in the work of Fourier: $$ \int_{a}^{b}f(x)g(x)w(x)\,dx. $$ In the 1830's, general Fourier problems with eigenvalue parameter $\lambda$, coefficient $p > 0$, and positive weight function $w$ were being studied by Sturm, Liouville and others: $$ -\frac{d}{dx}\left(p\frac{df}{dx}\right)+qf = \lambda wf,\\ \cos\alpha f(a)+\sin\alpha f'(a) = 0,\\ \cos\beta f(b)+\sin\beta f'(b) = 0. $$ It was found that there was an infinite sequence of real characteristic values $$ \lambda_{0} < \lambda_{1} < \lambda_{2} < \cdots $$ for which corresponding non-trivial solutions $\{ f_{n}\}$ would exist. And one would automatically have $$ \int_{a}^{b}f_{n}(x)f_{m}(x)w(x)dx = 0,\;\;\; n\ne m. $$ That allowed people to study generalized Fourier expansions in the eigenfunctions $\{ f_{n}\}_{n=0}^{\infty}$. This was an active area of research starting in the 1820's, especially the problem of when the corresponding Fourier expansions would converge to the original function.

So, many types integral "orthogonality" conditions were being studied in the first decades of the 19th century. Cauchy had proved what is now called the Cauchy-Schwarz inequality for Euclidean space in the 1830's, but, for some reason, no connection was made with the analogous integral expressions. It wasn't until the 1880's, in a paper of Schwarz, that a general so-called Cauchy-Schwarz inequality came into use for integrals, and that the paved the way for looking at general inner products in the early 20th century. The full abstraction of dealing with functions as points in a space, with orthogonality and distance, appears to be have been due to Hilbert in the first decade of the twentieth century. So, what started as a convenient type of integral 'orthogonality' condition for isolating coefficients in a generalized Fourier series eventually gave rise to general inner product spaces--the full abstraction took more than a century.

The particular choice of inner product has to do with the types of orthogonal expansions that you want to do.

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  • $\begingroup$ Great info--do you have a reference for more on the history you have described? $\endgroup$ – user45664 Jul 8 '16 at 20:58
  • $\begingroup$ @user45664 : J. Dieudonne History of Functional Analysis. amazon.com/… $\endgroup$ – DisintegratingByParts Jul 8 '16 at 21:29
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In the vector space $V$ of real valued functions $f\colon \mathbb R \to \mathbb R$ (some variations are possible, but the gist remains the same) the standard inner product is defined as you describe. It is the standard inner product, not the only inner product. The reason is that it is a natural generalization of the standard inner product on $\mathbb R^n$, which itself is a natural generalization of the standard inner product on $\mathbb R^2$, namely $(x_1,x_2)\cdot (y_1,y_2)=x_1y_1+x_2y_2$. So, the question boils down to why is that considered the standard inner product. This is where geometry comes in. The Law of Cosines is a generalization of Pythagoras theorem relating lengths and angles between vectors. In more detail, in a triangle of side lengths $a,b,c$, one has $c^2=a^2+b^2-2ab\cos(\gamma)$ where $\gamma$ is the angle opposite $c$.

Now, using Pythagors to define the length of a vector in $\mathbb R^2$ as usual, and then pluging things into the law of cosines yields, after some canceling, a relationship between the sum of the products of the components of the vectors, their lengths, and the cosine of the angle in what is precisely the Cauchy-Schwarz inequality. This motivates the definition of the standard inner product.

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    $\begingroup$ thanks, but u mean that the inner product of 2 functions is a generalization of the usual inner product for n-dimensional vectors as a function can be thought of as a vector with infinite dimensions? this means that $F_1(x).F_2(X)=\sum f_{1i}f_{2i}, and \int F_1(x)F_2(x)dx=\sum f_{1i}(x)f_{2i}(x) \Delta x$, there's a delta x missing $\endgroup$ – Mohamed M. Saad Feb 13 '15 at 1:26
  • $\begingroup$ in a nutshell, yes. $\endgroup$ – Ittay Weiss Feb 13 '15 at 1:28
  • $\begingroup$ yeah, but where's the $\Delta$ x? I thought of this before and this is what I reached, how is the product denoted by the integral when the summation doesn't have a "dx" to turn it to an integral? $\endgroup$ – Mohamed M. Saad Feb 13 '15 at 1:41

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