# Show boundedness for a linear functional defined on n linearly independent vectors

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 $$|l(y)|\leq C||y|| \ \forall y\in Y.$$

• 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 $b_i$. Commented Oct 29, 2023 at 20:19