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Given:

A = $\{ \vec{v} = (v_{1}, ....v_{k}, ...) \in \mathbb{R}^{\infty} | \sum_{i=1}^\infty v_i^2 \text{converges} \}$

Prove that the set is a subspace of $\mathbb{R}^\infty$ and: < $\vec{v}, \vec{u} > = \sum_{i=1}^\infty v_iu_i$ defines an inner product on A.

I am really struggling with this problem, particularly showing closure under addition to show the set is a subspace, and proving positive definiteness of the inner product. I have been given the hint by a professor "use the Cauchy-Schwarz inequality" but I lack confidence in this advice, it is my understanding that a well defined inner product is a pre-condition for using Cauchy-Schwarz inequality.

This is associated with an intro Linear Algebra take home exam, I have found a lot of information that leads me to believe this is regarding an $\mathscr{l}^2$ - Hilbert space, a concept that is not actually in my textbook.

How to go about this proof? Ideally in a way that someone with 1 semester of linear algebra could understand.

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    $\begingroup$ +1 for being up front and honest that this is an assignment. Hints: write out the sum of two elements in your vector space and break it into as many sums as you can over as many terms as you can. do you see what parts converge for sure? $\endgroup$ – The Count May 15 '17 at 1:14
  • $\begingroup$ Hint: the function $f(x)=x^2$ is convex. $\endgroup$ – tst May 15 '17 at 1:25
  • $\begingroup$ Wow.... It was too obvious how to "prove" that a sum of squares is necessarily positive! thank you $\endgroup$ – hulud May 15 '17 at 1:39
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Let $x,y \in A$ where $$x=(x_1,x_2....)$$ $$y=(y_1,y_2....)$$

Then $x+y=(x_1+y_1,x_2+y_2.....)$ and

$\sum_{n=1}^{\infty} (x_n+y_n)^2=\sum_{n=1}^{\infty}x_n^2+2x_ny_n+y_n^2 \leqslant \sum_{n=1}^{\infty}2x_n^2+2y_n^2=2\sum_{n=1}^{\infty}x_n^2+2\sum_{n=1}^{\infty}y_n^2 < \infty$ thus $x+y \in A$(i used the fact that $x^2+y^2 \geqslant 2xy \Longleftrightarrow(x-y)^2 \geqslant 0$)

If $a \in \mathbb{R}$ and $x=(x_1,x_2...) \in A$ then $\sum_{n=1}^{\infty}a^2x_n^2=a^2\sum_{n=1}^{\infty}x_n^2 < \infty$ thus $ax \in A$.

We proved that $A$ is a subbase of $\mathbb{R}^{\infty}$

Now $<x,x> $ where $x \in A$ is non negative because it is a sum of squares.

It is easy to prove the other axioms of inner product

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  • $\begingroup$ Thank you, this helps quite a bit. For positive definiteness though, how to show that $<\vec{x}, \vec{x}> = 0$ iff $\vec{x} = \vec{0}$ $\endgroup$ – hulud May 15 '17 at 1:38
  • $\begingroup$ this is obvious..if a sum of squares of real numbers is zero then all the real numbers in the sequence must be zero..i can state also in this way : $ \forall n \in \mathbb{N} ,0 \leqslant x_n^2 \leqslant \sum_n x_n^2=0$ thus $x_n^2 =0$ for every $n$ thus $x_n=0$ for every $n$ $\endgroup$ – Marios Gretsas May 15 '17 at 1:45
  • $\begingroup$ if $x=0$ then every component of $x$ is $0$ thus every component in the second power is $0$ thus the whole sum (which is equal to $<x,x>$) is $0$ $\endgroup$ – Marios Gretsas May 15 '17 at 2:10
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To check positive definiteness - clearly $\langle x,x\rangle\ge 0$ as it is a sum of squares. Suppose $x_i\neq 0$ for some $i$ - what does this tell you?

After you have check symmetry and linearity, you can apply Cauchy Schwarz since it is an inner product. Can you expand $\langle x+y,x+y\rangle$ using linearity, and apply Cauchy Schwarz to show it is finite?

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