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Consider $\{1, t, t^2, t^3, \ldots\}$ as a subset of $C\big( [0, 1]\big)$. Clearly this system is linearly independent. Also, it is a complete system (meaning that its closed linear hull is the whole space $C\big( [0, 1]\big)$ ) but it is not a Schauder basis (meaning that not any continuous function can be decomposed in a uniformly convergent sum $\sum_0^\infty a_kt^k$).

Lying somewhere in between those two notions is the concept of a minimal system: a linearly independent and complete system $\mathbf{x}=\{x_n\}_{n=1}^\infty$ of vectors of a Banach space $X$ is called minimal iff there exists a system $\mathbf{x}^\star=\{x_n^\star\}_{n=1}^\infty$ of bounded linear functionals such that $\langle x_n^\star, x_m\rangle=\delta_{n, m}$ (if this is the case, the system $\mathbf{x}^\star$ is called biorthogonal to $\mathbf{x}$). Equivalently, $\mathbf{x}$ is a minimal system iff the natural projections

$$P_m\left( \sum_{j=0}^na_jx_j\right)=\sum_{j=0}^{\min(n, m)}a_j x_j$$

are bounded.

Question. Is the monomial system $\{1, t, t^2, \ldots\}$ minimal in $C\big( [0, 1]\big)$?

I am induced to think that the answer is affirmative. This would make for an example of a minimal system that is not a Schauder basis. (Indeed, a minimal system is a Schauder basis precisely when the projections $P_m$ are uniformly bounded - cfr. Lindenstrauss-Tzafriri Classical Banach spaces vol.I, Prop. 1.a.3).

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up vote 2 down vote accepted

No, e.g. the projection $P_1$ is not bounded: there are polynomials $f$ of arbitrarily small norm with $f(0) = 0$ and $f'(0) = 1$. For a nice example, let $g(t)$ be a Maclaurin polynomial of the entire function $\text{sinc}(kt) = \dfrac{\sin(k t)}{kt}$ for $t \ne 0$, $\text{sinc}(0)=1$, such that $|g(t) - \text{sinc}(kt)| \le 1/k$ for $t \in [0,1]$, and take $f(t) = t g(t)$. Note that $f(0)=0$, $f'(0)=g(0)=1$, and $|f(t)| \le \frac{1}{k} + \frac{|\sin(kt)|}{k} \le \frac{2}{k}$ for $t \in (0,1]$.

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Very nice indeed. Thank you. – Giuseppe Negro Sep 10 '11 at 8:14

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