Let $l^2$ be the set of sequences $x = (x_n)_{n\in\mathbb{N}}$ ($x_n \in \mathbb{C}$) such that $\sum_{k\in\mathbb{N}} \left|x_k\right|^2 < \infty$, how can I prove that $l^2$ is a Hilbert space (with dot-product $\left(x,y\right) = \sum_{k\in\mathbb{N}} x_k\overline{y_k}$). This is a standard textbook exercise: apparently this is easy and, even to me, it seems self-evident. However, I don't know what to do with the infinite sum.
How to prove that square-summable sequences form a Hilbert space?
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5"Hilbert space" encompasses a number of properties. Which one is causing you trouble? – 2011-03-09
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0Well, for finite sums, I have no problems; but how do I know, for example, if $\overline{\sum_{k\in\mathbb{N}} a_k} = \sum_{k\in\mathbb{N}} \overline{a_k}$ (for showing that $\overline{(y,x)} = (x,y)$)? Moreover, I have trouble showing that every Cauchy sequences is convergent in $l^2$. – 2011-03-09
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0Use the fact that, by definition, $\sum_{k\in\mathbb{N}a_k} = \lim\limits_{n\to\infty}\sum_{k=1}^n a_k$. Work with the finite partial sums, and use the properties of limits to show the properties you want also holds for the series. – 2011-03-09
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0Great, that solves half my problem! – 2011-03-09
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1LaTeX tip "of the day": try using `\ell_p` for the sequence space $\ell_p$ – 2011-03-09
3 Answers
Show that your definition $\left(x,y\right) \equiv \sum_{k\in\mathbb{N}} x_k\overline{y_k}$ Satisfies all the inner product axioms.
Hint: axioms are
- conjugate symmetric
- linearity in one of the arguments
- Positive definite: $(u,u)\geq 0$ equality iff u=0
Now show that $F(x) = (x,z)$ defines a bounded linear functional $F$ on your inner product space $X$ of norm $||z||$. If the mapping $z\rightarrow F$ ( $X\rightarrow X'$ where X' is space of continuous linear functionals) is surjective, it must be a hilbert space.
EDIT OK, since you're worried about infinites let me share something I discovered recently in a book called the Cauchy Schwarz masterclass, the author calls it the principle of qualitative inference.
Given: $\sum_{k\in\mathbb{N}} a_k^2 \lt \infty$ and $\sum_{k\in\mathbb{N}} b_k^2 \lt \infty$ then it can be shown that $\sum_{k\in\mathbb{N}} a_k b_k \lt \infty$
For large k
$a_k b_k \leq \frac{1}{2} (a_k^2+b_k^2)$
$\Rightarrow \sum_{k\in\mathbb{N}} a_k b_k \leq \sum_{k\in\mathbb{N}} \frac{1}{2} (a_k^2+b_k^2)$
This already gives you a glimpse at the truth of the assertion.
(On a sidenote, you can go on to introduce $\hat{a}_k=\frac{a_k}{\sqrt{\sum_{k\in\mathbb{N}} a_k^2}}$ and derive the Cauchy inequality!)
Now do for conjugate symmetric what was done for real symmetric.
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0I can do that for any finite sequences, I know I am missing something here, but I am not sure why infinite sums are still "well-behaved" for what I need here. – 2011-03-09
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1@NoNameHere: How much do you know about infinite series? In any case: remember that the infinite series is defined to be the *limit* of the partial sums, so you can always write out the limit, operate in the *finite* partial sums, and then use the properties of the *limit* to get what you want. – 2011-03-09
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0It's been three years since I had my last analysis class: I'm not really surprised to have difficulty in functional analysis... – 2011-03-09
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0@noname see if my edit addresses your questions about infinite sums – 2011-03-09
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0@Approximist: with use of limits properties and of the Cauchy-Schwarz, my questions about series are answered. – 2011-03-09
You need to show that $\ell^2$ is a vector space, that the dot product satisfies the properties of inner product, and that the space is complete relative to the induced metric
The tricky bit in showing it is a vector space is showing that it is closed under sums; this is a consequence of the Hölder inequality.
Showing that the dot product is an inner product is straightforward: you may need to use the Hölder inequality or Cauchy-Schwarz inequality to show that $\sum x_k\overline{y_k}$ converges when both $(x_n)$ and $(y_n)$ are square summ
And then you have the issue of showing that $\ell^2$ is actually complete relative to the induced norm. This is pretty standard (in fact, one shows that the spaces $\ell_p$ are Banach spaces for all $p\geq 1$, hence complete, so $\ell^2$ is complete.
Added. From the comments, it seems your main difficulty is operating with the infinite sums and making sure they have the properties you want, similar to finite sums. (Let me commend you for your caution: it is indeed important not to simply assume that properties that hold for finite sums will transfer to infinite series). We use the fact that, by definition, $$\sum_{k=1}^{\infty} a_k = \lim_{n\to\infty}\sum_{k=1}^n a_n.$$ In the sense that the series on the left converges to the limit, if the limit exists.
So, for example, you have: \begin{align*} \overline{\langle (x\_n),(y\_n)\rangle} &= \overline{\sum\_{k=1}^{\infty}x\_k\overline{y\_k}}\\\ &=\overline{\lim\_{n\to\infty}\sum\_{k=1}^n x\_k\overline{y\_k}}\\\ &= \lim\_{n\to\infty}\overline{\sum\_{k=1}^n x\_k\overline{y\_k}}&&\mbox{(since conjugation is continuous)}\\\ &= \lim\_{n\to\infty}\sum\_{k=1}^n\overline{x\_k\overline{y\_k}} &&\mbox{(since the sum is finite)}\\\ &= \lim\_{n\to\infty}\sum\_{k=1}^n\overline{x\_k}\overline{\overline{y\_k}}\\\ &= \lim\_{n\to\infty}\sum\_{k=1}^n\overline{x\_k}y\_k\\\ &= \lim\_{n\to\infty}\sum\_{k=1}^ny\_k\overline{x\_k}\\\ &= \sum\_{k=1}^{\infty} y\_k\overline{x\_k}\\\ &= \langle (y\_k),(x\_k)\rangle. \end{align*} Etc.
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0The Cauchy-Schwartz inequality is the one fact I knew how (and when) to use (but before I had to show that the dot-product is well-defined, which I know see, is quite easy). I still struggle to show that the space is complete however. – 2011-03-09
This is more of an addendum for later, when you dig deeper into functional analysis:
Complex analysis tells us, that every holomorphic function can be represented by its Taylor series, locally. Actually, the space of square summable complex numbers is, as a Hilbert space, isomporph to all holomorphic functions on the unit disc $D := \{ z \in \mathbb{C}, |z| \le 1 \}$, that are square integrable: $$ \int_D |f(z)|^2 dz \le \infty $$ with the obvious scalar product $$ (f, g) = \int_D \bar f (z) g(z) dz $$ via the mapping $$ f(z) \to \sum_{i = 0}^{\infty} a_i z^i $$ and vice versa.
It may help to compare the proof of the Hilbert space axioms for the space of sequences with the proof of the axioms for the function space (the latter needs some "advanced" knowledge of complex calculus). Some may seem to be easier, some more involved. For example it is easy to see that the scalar product actually is a scalar product, without the need to handle infinite sums.
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0I might misunderstand what you're saying. While it is true that any 2 infinite dimensional separable Hilbert spaces are isometrically isomorphic, the space of holomorphic functions on the disk whose sequences of MacLaurin coefficients are in $\ell^2$ is the Hardy space $H^2$, which is not the space you described. You have described the Bergman space of the disk, $A^2$, which properly contains $H^2$, and in which $\|f\|_2^2=\pi\sum_{k=0}^\infty|a_k|^2/(k+1)$. – 2011-03-09
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0Will you please edit to clarify your post? You seem to be implying that the map $f\mapsto(a_0,a_1,\ldots)$ is an isomorphism from the Bergman space to $\ell^2$, when in fact for many $f$ in the Bergman space, $(a_0,a_1,\ldots)$ is not in $\ell^2$. That is, I think that your post is misleading for the same reason I commented earlier. – 2011-03-12