MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Is the dimension of the space of $H^2(B)$ harmonic functions on unit ball $B\subset\mathbb{R}^d$ countably or uncountably infinite?

share|cite|improve this question
It is a complete Banach space of infinite dimensions: such a thing is never countably dimensional. – Mariano Suárez-Alvarez Jul 5 '10 at 21:26
...unless you mean the Hilbert dimension of $H^2(B)$, which is countably infinite since it's a separable infinite dimensional Hilbert space. – Pietro Majer Jul 5 '10 at 22:28

Robin Chapman has given a complete answer, but I want to add something that is too long for a comment. The point is the answer depends on how you wish to define "dimension." Strictly speaking, the previous sentence is non-sense as there is no disagreement on how to define dimension: it is the cardinality of a maximal linearly independent subset. For the purpose of this answer, let's call this "algebraic dimension." My reason for giving it a special name is that I suspect that what you REALLY want to know is if this space is separable: "does it have a countable dense subset?" Let us call the cardinality of a minimal set with dense span the "analytic dimension." So a separable space is one with countable analytic dimension.

Notice that in order to find the analytic dimension you need to know what "dense" means. That is you need a topology, which you must define. However, the answer to your question is "yes" for most of the usual function space topologies you might put on your space. Actually that depends on what space you mean exactly. If you want to consider ALL harmonic functions, you probably want to use the topology of uniform convergence on compact sets. Your notation suggests maybe that you have in mind square integrable harmonic functions, for which you would naturally use the $L^2$ norm. (Note: you will run into trouble if you want to consider the space of bounded harmonic functions with the sup norm -- this space is not separable.)

As Robin explained very well, the algebraic dimension of your space is uncountable. In fact, from his answer you can glean that pretty much any interesting function space will have uncountable "algebraic dimension." On the other hand, many function spaces with their natural topologies are separable, i.e. have countable "analytic dimension." The big exceptions are spaces like $L^\infty$ that involve a sup norm and no regularity.

share|cite|improve this answer

Uncountably infinite as long as $d\ge2$.

We can solve the boundary value problem for countinuous functions on the unit sphere $S$. So we get a linear injection from $C(S)$ to $H(B)$. Now the dimension of $C(S)$ is uncountable. The unit interval $[0,1]$ embeds in $S$ and by Urysohn's lemma we get a linear surjection from $C(S)$ to $C([0,1])$. To see that $C[0,1]$ has uncountable dimension, note that the functions $f_t:x\mapsto e^{tx}$ are linearly independent for $t\in\mathbb{R}$.

Added A less clumsy way of proving that $C(S)$ is uncountably dimensional than the hasty argument above is to consider $F_t:(x_1,\ldots,x_n)\mapsto e^{t x_1}$.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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