I am trying to understand the proof of this theorem:

Let $(y_i)_1^n$ be $n$ linearly independent vectors in a normed linear space $X$ and $(a_i)_1^n$ be $n$ complex numbers. Then there exists a linear functional $l$ s.t. $$l(y_i)=a_i\ \forall i.$$

Proof: Let $Y$ be the linear space spanned by $(y_i)_1^n$; it consists of vectors of the form $$y=\sum_1^nb_iy_i.$$ Since the $y_i$ are linearly independent, the representation of $y$ is unique. Now define $l$ on $Y$ by $$l(y)=\sum_1^nb_ia_i.$$ Clearly, $l$ is linear and bounded on $Y$ and satisfies the required condition. It can then be extended to $X$ by Hahn-Banach.

The thing I'm uncertain of is boundedness. I know that boundedness and continuity (just at $0$ suffices) are equivalent for linear functionals and it's easy to show continuity at $0$, but I wanted to know if there's an easy way to obtain a bound of the style $$|l(y)|\leq C||y|| \ \forall y\in Y.$$

Thanks in advance

  • $\begingroup$ If you mean finding $C$ explicitly, I don't think there is a general method. The problem is that there does not need to be an easy way to express $\|y\|$ in terms of the $b_i$. $\endgroup$ Commented Oct 29, 2023 at 20:19


You must log in to answer this question.

Browse other questions tagged .