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I know that in infinite dimensional Hilbert spaces sometimes the best we can do is to find an orthonormal basis in the sense that any element in H can be approximated arbitrarily close in the NORM by a finite linear combination of this basis elements.

So then does that mean we can't expect that every x in H could be written as a finite linear combination of basis elements correct? So then we can't have things like $x= \sum_{i=1}^{\infty} a_k e_k $ for some $ a_k$ ARE constants in the underlying field, usually $\mathbb C$ (usually the projections of x on each $e_k$). So then how do we deal with linear transformations? For example how do we even define what a linear transformation does without explicitly saying what T(x) is for each x in H...ie how is saying what $T(e_k)$ is for each k enough to describe the whole linear transformation?

Thanks for answers to either question.. I see in a lot of proofs people writing x as this kind of infinite sum which confuses me since the infinite sum might not be in H..

Finally if we take some kind of infinite sum of elements in H, and the norm of that is finite, can we conclude the infinite sum is in H or that's still not enough?

  • Are you asking about finite sums or infinite sums? You start with "we can't expect that every x in H could be written as a finite linear combination of basis elements correct?" and then say So then we can't have things like $x= \sum_{i=1}^{\infty} a_k e_k $. But that sum is not a finite linear combination. – Ben Grossmann Apr 04 '17 at 18:05
  • Either one basically. –  Apr 04 '17 at 19:18
  • Ok. Well, you have two answers now, so let us know if either makes sense to you, or if you need clarification. – Ben Grossmann Apr 04 '17 at 19:29
  • Note that there is a significant difference between finite linear combinations and (allowing for) infinite linear combinations, hence the question. It seems like you're confusing the two. – Ben Grossmann Apr 04 '17 at 19:33

2 Answers2

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If $\{e_k\}$ is an orthonormal basis of your Hilbert space $H$, then every $x\in H$ can be expressed in the form:

\begin{equation} x=\sum_{k} a_k e_k \end{equation}

with, as you say, the coefficients $a_k$ in the field (most usually $\mathbb C$). It is important to note that the choice of the coefficients is unique, namely $a_k=<x,e_k>$.

The converse is not true in general: for a sum like the one above to be in $H$ it is necessary (and sufficient) that $$ \sum_{k} |a_k|^2<\infty. $$

Of course, that would be always true if the dimension of the space (the cardinality of $\{e_k\}$) is finite. Hence, the infinite dimensional case is the only one of actual interest in this context.

Now, if you have a linear transformation $T$ defined in each of the $e_k$ then that determines $T$ on the whole of $H$, provided that the transformation is bounded, that is when $$ T(e_k)\leq M $$ for some fixed constant $M>0$.

In this case there is no problem: $$ T(x)=\sum_{k} a_k T(e_k) $$ and the sum in the right hand side is guaranteed to converge in $H$. It is readily verified that in the finite dimensional case every linear transformation is bounded, so again the case of interest is when the space is infinite dimensional.

However, if $T$ is not bounded, the most you can expect to have is an operator defined in a dense linear subspace of $H$ (called the domain of the operator, and denoted $D(T)$): this domain is the set of the $x$ for which the sum above converges in $H$.

Federico
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  • You're completely ignoring the question about finite linear combinations. At the very least, you should be explicit whether or not your sums are infinite. – Ben Grossmann Apr 04 '17 at 18:08
  • As far as I can see, everything stated above is true irrespective whether the space is finite or infinite dimensional. And in particular the sums considered can be both finite or infinite. Please, point out where you think this needs clarification. Thanks. – Federico Apr 04 '17 at 18:28
  • For instance: if we're only talking about finite sums, then there's no need for $T$ to be a bounded operator. If we're talking about sums that could potentially be infinite, then we're effectively talking about all infinite sums anyway. – Ben Grossmann Apr 04 '17 at 19:31
  • Ok, I added a couple of lines in order to address the difference between the infinite and finite dimensional cases (more precisely, that the finite dimensional case is covered, as a trivial particular case for the questions at hand). – Federico Apr 05 '17 at 17:47
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There are two notions of a basis. One is the idea of a Hamel basis, and the other is the idea of a Schauder basis. For more on that, see this post. Every vector space has a Hamel basis. However, in a Hilbert space, one primarily talks in terms of Schauder bases.

So, every element in $H$ can be written as an "infinite linear combination" of elements from our orthonormal basis. Moreover, because our basis is orthonormal, the sum $\sum_{k=1}^\infty a_ke_k$ is convergent in $H$ if and only if $\sum_{k=1}^\infty |a_k|^2$ converges, and we have $\left\| \sum_{k=1}^\infty a_ke_k \right\|^2 = \sum_{k=1}^\infty |a_k|^2$.

Remember that any infinite sum is really a limit. In particular, $\sum_{k=1}^\infty a_ke_k = \lim_{N \to \infty}\sum_{k=1}^N a_k e_k$. This limit is defined with respect to the norm on a Hilbert space. The limit exists whenever $\sum_{k=1}^\infty |a_k|^2$ is finite because Hilbert spaces are complete.

Whenever $T$ is a continuous (equivalently, bounded) linear operator, it suffices to define $T$ over our orthonormal basis. Because $T$ is continuous, we can say $$ T(\sum_{k=1}^\infty a_ke_k) = T(\lim_{N \to \infty}\sum_{k=1}^N a_ke_k) = \lim_{N \to \infty}T(\sum_{k=1}^N a_ke_k) = \lim_{N \to \infty}\sum_{k=1}^N a_kT(e_k) $$ In a sense, we can still use this trick when $T$ is a closed operator, but we need to remember that $T$ is only defined over its domain. The closedness tells us that whenever the sum on the right converges, it coincides with the actual output.

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