Show that a set is complete in a generic Hilbert space $H$

Let $\{e_n\}$, $n\geq1$ be an orthonormal base of an hilbert space $H$. Let $$f_1=e_1-e_2+e_3,\quad f_2=e_2-e_3+e_4,\quad \ldots\quad f_n=e_n-e_{n+1}+e_{n+2}.$$ Show that $\{f_n\}$, $n\geq1$ is complete. I think I have to use the definition, that is $$(f_n, x)=0,\ \forall n\ \ \Rightarrow \ \ x = 0$$ but I cannot understand how. Some help?

• What does it mean for a sequence of vectors to be complete? – Matthew Leingang Sep 26 '17 at 13:33
• what do you mean by "generic?" – supinf Sep 26 '17 at 13:33
• @supinf an hilbert space $H$. The exercise doesn't specify.. – Jeji Sep 26 '17 at 13:35
• @MatthewLeingang a definition is included, but it looks more like dense instead of complete – supinf Sep 26 '17 at 13:35

Assume $\langle x, f_n \rangle = 0$, for every $n \in \mathbb{N}$. Notice that for $n \ge 2$ we have:

$$\sum_{k=1}^n f_k = \sum_{k=1}^n e_k-e_{k+1}+e_{k+2} = e_1+e_{n+2}$$

So

$$0 = \sum_{k=1}^n\left\langle x, f_k\right\rangle = \left\langle x, \sum_{k=1}^n f_k\right\rangle = \langle x, e_1 \rangle + \langle x, e_{n+2} \rangle$$

which implies

$$\langle x, e_1 \rangle =- \langle x, e_{n} \rangle, \quad \forall n \ge 4$$

Now, since $(e_n)_{n=1}^\infty$ is an orthonormal basis we have:

$$\|x\|^2 = \sum_{n=1}^\infty \left|\langle x,e_n\rangle\right|^2 = \left|\langle x,e_1\rangle\right|^2 + \left|\langle x,e_2\rangle\right|^2 + \left|\langle x,e_3\rangle\right|^2 + \sum_{n=4}^\infty \left|\langle x,e_1\rangle\right|^2$$

For the sum to be finite, it must hold $\langle x, e_1 \rangle = 0$.

Now

$$0 = \langle x,f_3\rangle = \langle x,e_3-e_4+e_5\rangle = \langle x,e_3\rangle - \langle x,e_4\rangle + \langle x,e_5\rangle = \langle x,e_3\rangle$$

so we have $\langle x,e_3\rangle = 0$.

Finally:

$$0 = \langle x,f_1\rangle = \langle x,e_1-e_2+e_3\rangle = \langle x,e_1\rangle - \langle x,e_2\rangle + \langle x,e_3\rangle = -\langle x,e_2\rangle$$

So, $\langle x,e_2\rangle = 0$.

Therefore, we have $\langle x,e_n\rangle = 0$, $\forall n\in\mathbb{N}$.

Now, since $(e_n)_{n=1}^\infty$ is an orthonormal basis:

$$x = \sum_{n=1}^\infty \underbrace{\langle x,e_n}_{=0}\rangle e_n = 0$$

An orthonormal set $\{f_i\}_{i\in I}$ with this property is sometimes called a maximal orthonormal set, and in a Hilbert space this is equivalent to $\{f_i\}_{i\in I}$ being an orthonormal basis. It is maximal in the sense that there does not exist a vector $f \in H$, such that $f \perp f_i, \forall i\in I$ and that $\{f_i\}_{i\in I} \cup \{f\}$ is also an orthonormal set.

• Possible short-cut: Let $H_n$ be the closed linear subspace of $H$ generated by $\{e_j:j\geq n\}$. Let $F_n$ be the closed linear subspace of $H_n$ generated by $\{f_j:j\geq n\}.$ Your method for showing for $x\in H_1$ that $x\in F_1^{\perp}\implies <x,e_1>=0\implies x\in H_2$ can be applied for all $n:$ If $x\in H_n$ and $<x,y>=0$ for all $y\in F_n$ then $x\in H_{n+1}.$ By induction on $n,$ if $x\in F_1^{\perp}$ then $x\in H_n$ for all $n,$ so $x=0.$.......................+1 – DanielWainfleet Sep 26 '17 at 21:47
• $\{f_i:i\in I\}$ is not orthonormal . $<f_1,f_2>=-2.$ – DanielWainfleet Sep 26 '17 at 21:58

Hint: By using $f_1+f_2=e_1+e_4$ we get $$0= (f_1,x) + (f_2,x) = (e_1,x)+(e_4,x)$$

Doing the same with $f_4,f_5$ and combining with the above, yields $$(e_1,x)=(e_7,x).$$

This can be generalized, so you have statements about $(e_n,x)$ for $n\in \mathbb N$.

Then you have to use $(e_n,x)\to0$, which follows from $e_n \rightharpoonup 0$ or Bessels inequality.

After that you will be much closer to the goal.

• edited to adress your questions – supinf Sep 26 '17 at 14:21
• after that, you know the value for each $(e_n,x)$. – supinf Sep 26 '17 at 14:59
• It is important that the property you are after is not the same as completeness. supinf has already noted this in their answer. A more appropriate word for the property you are after is that the set generates the Hilbert space. – s.harp Sep 26 '17 at 16:46