# Bound on projection to finite dimensional subspace of Banach Space

Let $M$ be a finite dimensional subspace of a Banach space $X$ with basis $\{x_1,\dots,x_n\}$ (through scaling we can assume each basis element has unit norm). We can construct a projection $P:X\to M$ as follows. For $i\in \{1,\dots ,n\}$ consider the linear coefficient functional $f_i':M\to \mathbb{F}$ defined on the basis elements by $f_i'(x_k)=\delta^i_k$. Now from Lemma 2.4-1 [1] we know that there exists a $c>0$ such that for any $i\in\{1,\dots,\}$ and $x=\sum_{k=1}^n\alpha_k x_k\in M$ we have that $$|f_i'(x)|=|\alpha_k|\leq \sum_{k=1}^n|\alpha_k|\leq c\|x\|_M\quad(1).$$ Hahn-Banach allows us to extend each $f'_i$ to $f_i:X\to \mathbb F$, where $f_i$ has all the regular properties of an extended functional. We can now define $Px=\sum_{k=1}^nf_k(x)x_k$ for each $x\in X$. It is easy to see that $P$ is linear, idempotent, and from $(1)$ that $\|P\|\leq cn$.

I would like to know whether there is any way to improve this bound? I think critically I would like to know more about the behaviour of the constant $c$. Kreyzsig uses proof by contradiction to prove the existence of the $c$, and I have not been able to find another direct proof so I can figure out a way to bound $c$. My intuition says that $c$ should be bounded, perhaps even equal $1$, as I've now spent some time being unable to come up with a counter-example. However, I have been unable to to come up with any bound either. I've played around with trying to use Riesz's Lemma, but without success. Any ideas and/or references would be greatly appreciated.

References:

[1] Erwin Kreyzsig. Introductory Functional Analysis with Applications. Wiley, 1st edition, 1989.

• @TomekKania Sorry I was busy with other work, and only recently had time to study the Kadets-Snobar theorem, and had forgotten about this post. This is in fact far more than I need, so thank you for the reference. – K.Power Apr 28 '17 at 22:50
• @TomekKania I have another, different, question about whether there's a bound for the $c$ in the theorem I quoted above. It's not in relation to the bound of the projection operator, which is answered by your answer, but I am curious. Don't know whether I should open a new question for it though. – K.Power Apr 28 '17 at 22:52

Yes, every $n$-dimensional subspace of a normed space is $\sqrt{n}$-complemented; this is the famous Kadets-Snobar theorem. By a result of Pisier, this is asymptotically optimal up to a multiplicative constant.