Generalized Monomials Let $P$ be the set of all power functions $p_a$: $(0, \infty)\longrightarrow\mathbb{R}$, $x\mapsto x^a$, indexed by $a\in [0, \infty)$.
Is $P$ linearly independent over $\mathbb{R}$? Intuitively, the answer is clearly yes (at least to me), but it seems to be escaping straightforward proof attempts.
Follow-up question:
Does the subspace of $C^{\infty}((0, \infty))$ spanned by $P$ admit some "nice" characterization?
 A: The proof of linear independence is the same as for integer powers. Let $0\le a_1<a_2<\dots<a_n$ and suppose that
$$
\sum_{k=1}^n\lambda_k\,x^{a_k}=0,\quad x\in\mathbb{R}.
$$
Divide by $x^{a_1}$, take derivatives and multiply by $x$ to obtain
$$
\sum_{k=2}^n\lambda_k(a_k-a_1)\,x^{a_k-a_1}=0,\quad x\in\mathbb{R}.
$$
Repeat $n-1$ times to get $\lambda_n=0$.
I do not think there is a characterization of the algebraic spann of $P$. You could call them generalized polynomials.
A: $P$ is indeed linearly independent over $\mathbb R$, since the exponential is bijective since invertible (logarithm) which gives pairwise different values for $x > 0$ in any finite collection of monomials. Also, the functions grow the quicker the higher the value of $a$ (at least for $x > 1$). This allows for a construction of linearly independent vectors of the form
$$
\begin{pmatrix}
m_1(x_1) \\ m_1(x_2) \\ \vdots \\ m_1(x_n)
\end{pmatrix}, \begin{pmatrix}
m_2(x_1) \\ m_2(x_2) \\ \vdots \\ m_2(x_n)
\end{pmatrix}, \ldots, \begin{pmatrix}
m_n(x_1) \\ m_n(x_2) \\ \vdots \\ m_n(x_n)
\end{pmatrix},
$$
which must be linearly dependent if $m_1, \ldots, m_n$ are. The linear independence of those vectors can be proven by arguing why they can be reduced to the identity by elementary matrix operations; normalizing the first row, then clearing out everything below the pivot in the first column and then arguing why a smaller matrix of the same form is left. It is convenient if the monomials are ordered by the value of $a$, and the $x_j$ are ordered by size.
They can't span $C^\infty((0, \infty))$; not even countable linear combinations would suffice, since the bump function is zero after a certain point, and any nontrivial countable sum from your space would not be zero on a whole interval, for it is positive.
