Given an orthonormal basis $(e_n)_{n\in\mathbb N}$ of $H_0^1(\Lambda)$, construct an orthonormal basis of $H_0^1(\Lambda)^d$

Question

Let

• $k,d\in\mathbb N$
• $\Lambda\subseteq\mathbb R^k$ be open
• $(e_n)_{n\in\mathbb N}$ be an orthonormal basis of $H_0^1(\Lambda)$ and $$f^n:=\frac1d(\underbrace{e_n,\ldots,e_n}_{d\text{-times}})\;\;\;\text{for }n\in\mathbb N$$

Can we conclude that $(f^n)_{n\in\mathbb N}$ is an orthonormal basis of $H_0^1(\Lambda)^d$? That should be the case, cause $$\langle f^i,f^j\rangle_{H^1(\Lambda,\:\mathbb R^d)}=\langle e_i,e_j\rangle_{H^1(\Lambda)}\;\;\;\text{for all }i,j\in\mathbb N\;.$$

Answer 1

It is not true even in finite dimensional space. For example, if $V$ is $n$-dimensional, $V^d$ is $nd$-dimensional, so an orthonormal basis should have $nd$-elements instead of $n$ elements.

Instead, try

$$f^k_n := (0,0,\cdots 0, e_n,0,\cdots, 0),$$

(the nonvanishing term is at the $k$-th entry). Then $$\{ f^k_n\}_{k=1,\cdots,d, n\in \mathbb N}$$

forms an orthonormal basis for the direct sum. Note that this is true for all Hilbert spaces. Related question has been asked here