Orthonormal basis $L^2(a,a+2\pi)$

Question

Let $$\mathcal{B}=\left \{\frac{1}{\sqrt{2\pi}},\frac{\cos x}{\sqrt{\pi}},\frac{\sin x}{\sqrt{\pi}},\frac{\cos 2x}{\sqrt{\pi}},\frac{\sin 2x}{\sqrt{\pi}},\dots\right \}$$. This is an orthonormal basis of $L^2(a,a+2\pi)$ since its elements are orthonormal and $\overline{\operatorname{span}_{\mathbb{R}}B}=L^2(a,a+2\pi)$. Is it true that these vectors are also linearly independent?

Answer 1

yes, being orthonormal implies being linearly independent. assume $$ v_1, v_2, \ldots v_n$$ are orthonormal and suppose, without loss of generality that $$v_n = \sum_1^{n-1} a_i v_i$$ then $$ ||v_n||^2 = (v_n,v_n) = (\sum_1^{n-1} a_i v_i, v_n ) = \sum_1^{n-1} a_i (v_i, v_n) = 0 $$ which is a contradiction