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?

  • 1
    $\begingroup$ What does it mean for a sequence of vectors to be complete? $\endgroup$ – Matthew Leingang Sep 26 '17 at 13:33
  • $\begingroup$ what do you mean by "generic?" $\endgroup$ – supinf Sep 26 '17 at 13:33
  • $\begingroup$ @supinf an hilbert space $H$. The exercise doesn't specify.. $\endgroup$ – Jeji Sep 26 '17 at 13:35
  • $\begingroup$ @MatthewLeingang a definition is included, but it looks more like dense instead of complete $\endgroup$ – 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:

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


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

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 $\langle x, e_1 \rangle = 0$.


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

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


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

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.

  • 1
    $\begingroup$ 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 $\endgroup$ – DanielWainfleet Sep 26 '17 at 21:47
  • $\begingroup$ $\{f_i:i\in I\}$ is not orthonormal . $<f_1,f_2>=-2.$ $\endgroup$ – 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.

  • $\begingroup$ edited to adress your questions $\endgroup$ – supinf Sep 26 '17 at 14:21
  • $\begingroup$ after that, you know the value for each $(e_n,x)$. $\endgroup$ – supinf Sep 26 '17 at 14:59
  • 1
    $\begingroup$ 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. $\endgroup$ – s.harp Sep 26 '17 at 16:46

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.