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To prove that two objects are equal show that in enough circumstances they behave in the same way

Quick description

Suppose that you have two objects and would like to prove that they are in fact the same. If neither object has many degrees of freedom, in some appropriate sense that depends on the problem, then you can often establish this equality by performing a small number of tests on the two objects and showing that you get the same answers.


Basic linear algebra.

Note iconIncomplete This article is incomplete. More (and better) examples needed. One of these should be an application of the fact that if two polynomials of degree d agree in d+1 places then they are the same polynomial.

Example 1

A basic but important result in the theory of Hilbert spaces is that if e_1,e_2,\dots is an orthonormal basis, then v=\sum_{n=1}^\infty\langle v,e_n\rangle e_n for every vector v. How do we prove this? One way of doing so is to observe that both sides of this equation have the same inner product with any given e_m. Indeed,

\langle\sum_{n=1}^\infty\langle v,e_n\rangle e_n,e_m\rangle=\sum_{n=1}^\infty\delta_{nm}\langle v,e_n\rangle=\langle v,e_m\rangle.

To complete the proof, we need to know that any two vectors that have the same inner product with every e_m must be equal. (The main theme of this article is on using this sort of principle, however.) To prove this, it is enough to prove that if x is any non-zero vector then there must be some m such that \langle x,e_m\rangle\ne 0. But we know that such an x can be written in the form \sum_{n=1}^\infty\lambda_ne_n, so all we have to do is pick some m such that \lambda_m\ne 0 and take the inner product with e_m.

Example 2

This is in fact a special case of the previous example. The Fourier inversion formula tells us that if f is a suitably nice function defined on [0,2\pi] then we have the equality f(x)=\sum_{n=-\infty}^\infty\hat{f}(n)e^{inx} almost everywhere, where \hat{f}(n)=\int_0^{2\pi}f(t)e^{-int}dt.

More precisely, the functions defined on both sides of this equation are equal in the space L_2[0,2\pi]. To prove this, one first establishes that the functions e^{int} form an orthonormal basis, and then one applies the general idea of the previous example.

A different but equivalent way of thinking about it is to say that if we know that f(x)=\sum_n\lambda_ne^{int}, then we can calculate \lambda_m by taking the inner product of both sides with the function e^{-imt}.

General discussion


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