If $F$ is an orthonormal basis ,we represent every $x \in X$ as a linear combination consists of finitely many terms ?

Question

If $F$ is an orthonormal basis in an inner product space $X$,can we represent every $x \in X$ as a linear combination consists of finitely many terms ?

in Erwin Kreyszig's book says no, but i dont understand why the answer is no. if $F$ was orthogonal or not, what could we say about this ? Thank u for your help.

Answer 1

Think to the space of odd, square-integrable functions over $(-\pi,\pi)$, equipped with the inner product: $$ \langle f,g\rangle = \int_{-\pi}^{\pi}f(x)\,g(x)\,dx. $$ The sequence given by: $$ f_n(x) = \frac{1}{\sqrt{\pi}}\sin(nx),\quad n\geq 1 $$ gives an orthonormal base, but if we consider any discontinuous function in our space, it cannot be written as a finite combination of the $f_n$s.

Answer 2

An orthonormal basis (or Hilbert basis) in an infinite-dimensional Hilbert space $H$ is not a basis in the algebraic sense. It is an algebraic basis of a subspace that is dense in $H$.