Is {sin((pi)n*x/a)} a Basis for L^2(0,a)?

[mplltt]

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I have a question that pertains to the Sturm-Liouville theory. Prove that {sin((pi)nx/a)} n=1 (a>0) is the basis for L2 (0,a).

 

[benorin]

A basis set for L^2 (0,a)

I will assume we are dealing with continuous, real-valued functions here, as in your posted theorem. A set of functions, say $$\left\{ f_{n}(x)\right\}_{n=1}^{\infty},$$ forms a basis for L2(0,a) if and only if

for $$g(x)\in L^2(0,a)$$ we have $$\left< f_{n}(x),g(x)\right> =0$$ for every $$n=1,2,\ldots$$ implies that $$g(x)=0,$$

where <,> denotes the inner product on $$L^2(0,a)$$ defined by $$\left< g(x) , h(x)\right> = \int_{0}^{a}g(x)h(x) dx$$

In the above requirement for a set of functions to be basis, it should be understood that

$$\left< f_{n}(x),g(x)\right> =0$$ for every $$n=0,1,2,\ldots$$

means that

$$\left< f_{0}(x),g(x)\right> =\left< f_{1}(x),g(x)\right> =\cdots =0$$

or equivalently that g(x) is orthogonal to every f_n(x),

the "implies that $$g(x)=0,$$" part means that and the only function g(x) for which this condition may be satisfied (if the set of functions be a basis) is the zero function.

Now since we are dealing with continuous, real-valued functions, the Stone-Weierstrass theorem (in particular, Weierstrass approximation theorem) applies and any $$g(x)\in L^2(0,a)$$ can be approximated by a polynomial with arbitrary precision, hence we will test our set of functions against $$g(x)=c_0+c_1x+c_2x^2+\cdots +c_mx^m=\sum_{k=0}^{m}c_kx^k$$ for $$m=0,1,2,\ldots$$, where not every c_k is zero. Here it goes:

$$\left< f_{n}(x) , g(x)\right> = \int_{0}^{a}\left( \sum_{k=0}^{m}c_kx^k \right) \sin\left( \frac{\pi nx}{a} \right) dx = \sum_{k=0}^{m}c_k\int_{0}^{a}x^k \sin\left( \frac{\pi nx}{a} \right) dx$$

then prove that if the last integral is zero for every n=1,2,..., then $$c_k=0$$ for k=0,1,2,...m and you're done. You might try using $$\sin (x) = \frac{e^{ix}-e^{-ix}}{2i}$$ to evaluate the integral.