# Projections onto $n$-codimensional subspaces of a Banach space: norms.

Hello, I'd like some help to find an answer I've been looking for since this morning. Let $X$ be a Banach space and let $Y$ be an $n$-codimensional subspace of $X$. Let $P$ be a projection from $X$ onto $Y$. Which is the best estimate for the norm of $P$? I found this information in an article by Bohnenblust as far as $n=1$ is concerned (that is, there always exists a projection $P$ such that $\|P\|\leq 2+\varepsilon$), but nothing satisfactory when the codimension increases. Thank you.

-

In many books$^*$ you can find the result that there is a projection of norm at most $\sqrt{n}$ onto any $n$ dimensional subspace of a Banach space. For reflexive spaces, this gives immediately that every $n$ codimensional subspace is the range of a projection that has norm at most $\sqrt{n} +1$. For non reflexive spaces, by using the principle of local reflexivity (which also is in many books), you get for any $\epsilon > 0$ the estimate $\sqrt{n} +1 + \epsilon$.  $*$ See, for example, Albiac and Kalton, Topics in Banach space theory", Theorem 12.1.6. In this book you can also find the principle of local reflexivity.
If you want a slightly better estimate, see  König, H., Tomczak-Jaegermann, N.: Norms of minimal projections. J. Func- tional Anal. 119 (1994), 253-280.  renyi.hu/~makai/project.pdf –  Bill Johnson Oct 15 '12 at 15:59
The obvious answer is: $2^n +\varepsilon$. Just iterate the Bohnenblust construction to $X\supset Y_{n-1}\supset \dots \supset Y_1 \supset Y_0=Y$ where $y_1,\dots ,y_n \in X$ span a complement of $Y$ and $Y_i$ is the span of $Y$ and $y_1,\dots,y_i$.
In a beginning course I often raise the problem of getting some estimate for the projection onto an $n$ dimensional (or $n$ codimensional) subspace, and reasonable students come up with something like this. Occasionally a really good one figures out how to get an estimate that is polynomial in $n$. Those who work on the problem appreciate the depth of the (relatively simple) Kadec-Snobar theorem I mentioned in my answer when later in the course I prove it. –  Bill Johnson Oct 15 '12 at 15:41