# Is there a basis for the continuous functions space?

I've been searching all over the Internet for this but without finding a satisfying answer. This might be a dumb question, but I would like to know the answer anyway.

Is there a set of continuous functions which when combined linearly (or not maybe) span all the functions space ? Could we decompose a log, a sine or an exponent to simpler components ? And if not why ?

I know that Fourier analysis is a powerful tool for functions decomposition, but I wanted to know If we could go further and decompose even trigonometric functions. I wondered if there was is theory about this ?

• Yes. In fact, every vector space has a Hamel Basis. For the space in question, see this. – David Mitra Oct 18 '14 at 21:55
• Expanding upon what David said, the set of all continuous maps that take on real (or complex) values forms a vector space, and using the Axiom of Choice shows that this vector space has a basis. It will be an extremely large basis, however, certainly uncountable. – Hayden Oct 18 '14 at 21:57
• Well, every vector space has a basis (which is implied, and in fact equivalent, to the axiom of choice), so yes, a basis exists. As for constructing such a thing... well, it'd be uncountable, so that could be tricky. – Milo Brandt Oct 18 '14 at 21:58
• @DavidMitra thanks for the link, very interesting ! – vphenix Oct 18 '14 at 23:47
• @Hayden And if you didn't have the axiom of choice, then the answer at math.stackexchange.com/a/151186/26369 would explain why the continuous functions might not have a Hamel basis. – Mark S. Oct 28 '16 at 17:43

We can construct a simple basis for all functions as follows. Let $f_r(x)=\left\{ \begin{array}{cl} 1 & \text{ if } x=r \\ 0 & \text{ otherwise} \end{array} \right.$. Then any function can be made of a linear combination of these very simple functions. However, this basis is not very useful or interesting.