Question

(i) Let V be a Banach space and let L ⊆ V be a linear manifold. Show that L, the closure of L, is a subspace of V .

(ii) Let H be a Hilbert space. Let B be an orthonormal set in H. Show that span B is dense in H if and only if B is an orthonormal basis.

Answer 1

CORRECTED SOLUTION

1. Let L ⊆ V be a linear manifold, and $\bar L$ be its closure. For every $x\in \bar L$ there exist a sequence $\{x_n\}_{n=1}^{\infty}\subset L$ which is converges to x.

Let $a, b\in\bar L$ are arbitrary vectors, $a=\lim_{n\to+\infty}a_n$, $b=\lim_{n\to+\infty}b_n$ with $a_n, b_n\in L$ for every n. Then $a_n+b_n\in L$ and $\lim_{n\to+\infty}(a_n+b_n)=a+b$. Hence, $a+ b\in\bar L$ .

Let $\lambda$ be any scalar constant, then $\lambda a_n\in L$ for every n and $\lim_{n\to+\infty}\lambda a_n=\lambda\lim_{n\to+\infty} a_n =\lambda a$.

Hence, $\lambda a\in\bar L$ .

Since $\bar L$ is a closer of L, it is shown that $\bar L$ is a closed linear subspace of V.

2. if B is an orthonormal basis of H, then every element x in H may be approximated by a finite linear combinations of vectors from B, which are form span B. Hence, span B is dense.

Conversely, let span B is dense in H. Since B is an orthonormal set, for any $x\ne y \in B |x-y|=\sqrt 2$ . Since H is separable, this means that B must be countable or finite. Let $B = \{v_1, v_2, ...\}$ and x be an element of H. We show that the series $x_n=\sum_{k=1}^n\langle x, v_k\rangle v_k$ converges to x.

Let $\varepsilon>0$ and $|x-\sum_{k=1}^n\lambda_kv_k|<\varepsilon$ (it is possible to find such linear combination, since span B is dense in H). Then $\langle x-x_n,v_k\rangle=\langle x,v_k\rangle-\langle x_n,v_k\rangle=\langle x,v_k\rangle-\langle x,v_k\rangle=0$, hence $\langle x-x_n,x_n-\sum_{k=1}^n\lambda_kv_k\rangle=0$ and, by Pythagorean theorem, $|x-\sum_{k=1}^n\lambda_kv_k|^2=|(x-x_n)+(x_n-\sum_{k=1}^n\lambda_kv_k)|^2 = |x-x_n|^2+|x_n-\sum_{k=1}^n\lambda_kv_k|^2$ , which implies $|x-x_n|^2=|x-\sum_{k=1}^n\lambda_kv_k|^2-|x_n-\sum_{k=1}^n\lambda_kv_k|^2<\varepsilon^2$ and $|x-x_n|<\varepsilon$.

For instance, if we take $\sum_{k=1}^n\lambda_kv_k=x_{n-1}$ , then we have $|x-x_n|^2=|x-x_{n-1}|^2-|x_n-x_{n-1}|^2\leq |x-x_{n-1}|^2$ which implies $|x-x_n|\leq|x-x_{n-1}|$.

Therefore, the sequense $|x-x_n|$ is monotonous, non-negative and starting from some n is lesser than $\varepsilon$. This means $x_n$ converges to x and, therefore, B is an orthonormal basis in H.