Consider the vector space $C[0,1]$ of continuous functions on $[0,1]$. The map
$$\langle f,g\rangle=\int_0^1 fg $$
defines an inner product on $C[0,1]$, with property that $\langle f,f\rangle\ge 0$ for all $f$, with equality only if $f=0$.
Fact 1. The set $\{1,x,x^2,\ldots\}$ of polynomial functions is linearly independent.
Fact 2. By applying Gram-Schmidt process, we can obtain another (ordered) countable set
$$
1=g_1, \,\, g_2, \,\, g_3, \,\, \ldots
$$
of polynomials ($g_n$ of degree $n$), which are orthonormal.
Fact 3. Given $f\in C[0,1]$, since
$$
\int_0^1 x^nf(x)dx=0 \hskip5mm \mbox{ for all } n\ge 1 \hskip5mm \Rightarrow \hskip5mm f\equiv 0 \mbox{ (easy to check)}
$$
so this means that a continuous function on $[0,1]$ orthogonal to $1,x,x^2,\ldots$ must be $0$ only; hence the continuous function orthogonal to $1,g_2,g_3,\ldots$ must also be $0$. Thus, the set $\{1,g_2,g_3,\ldots, \}$ is a maximal orthonormal set (and is countable).
Fact 4. $C[0,1]$ is of uncountable dimension over $\mathbb{R}$.
We get, from above facts that, $C[0,1]$ with above inner product, has no basis containing orthonormal vectors.
Question: (My question can be trivial, but I did not find any direction from references of Roman's Advanced Linear Algebra or Simmons' Topology and Modern Analysis. The question came to mind from this example, and some paragraphs in these two books. ) Is there any example of uncountable dimensional real inner product space, which possesses a basis of orthonormal vectors? Or it is always the case that an orthonormal set in any real inner product space should be finite or countable?
Best Answer
Yes. Let $X$ be a vector space of uncountable dimension and let $\{x_i : i\in I\}$ be a Hamel basis for $X$. For $x,y\in X$ there exist unique representations $x = \sum_{i\in I_1}c_ix_i$ and $y = \sum_{i\in I_2}d_ix_i$, where $I_1$ and $I_2$ are finite subsets of $I$ and $c_i,d_i$ are scalars. Define $$ (x,y) := \sum_{i_1\in I_1}\sum_{i_2\in I_2}c_{i_1}d_{i_2}\delta_{ij}. $$ Then $\{x_i : i\in I\}$ is an orthonormal set in $X$ which is also a Hamel basis for $X$.