# The vector space of all functions of type $f:\mathbb{R} \rightarrow \mathbb{R}$

I am looking to describe all functions of type $f:\mathbb{R} \rightarrow \mathbb{R}$ as a vector space. Is this possible? What is a basis? Can I write any function of this type as $$g(x) = \int^{\infty}_{-\infty}a(k)sin(kx) + b(k)cos(kx)dk$$ for $a(k), b(k) \in \mathbb{R}$, and $k, \in \mathbb{R}$. I am assuming that $sin(kx)$ is a basis for this function space.

Let's suppose we have a basis of cardinality $Card(\mathbb{R})$, call it $B = \{b_1(kx), b_2(kx), \ldots \}, k \in \mathbb{R}$.

I don't understand why people are saying that the vector is a finite linear combination, but I can write a general function as

$$g(x) = \int^{\infty}_{-\infty}a_1(k)b_1(kx)+ a_2(k)b_2(kx) + \ldots dk$$

This seems like a continuous combination. Can someone explain this?

• Calling that integral $f(x)$, what is $f(-x)$? Can you write $\cos x$ as some integral of that type? – AHusain Aug 8 '18 at 16:21

You could also define the vector space that is spanned by the functions $\sin (kx)$. That is a fine vector space but it is missing many of the functions from $\Bbb R$ to $\Bbb R$. Because we only allow finite sums of basis elements, all the functions in this space are continuous.
It depends on how you define “to describe”. Saying that it is the set of all functions from $\mathbb R$ into itself is a perfectly fine description. But don't expect to find an easy-to-describe basis of that space. And, no, the functions $x\mapsto\sin(kx)$ don't form a basis of this space. For instance, all those functions are odd, and therefore no even function (other than the null function) can be expressed as a linear combination of some of them.