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Consider a proper non-trivial subspace $M$ of a finite-dimensional normed space $X$. We can always find a basis $\mathcal B=\mathcal B_1 \cup \mathcal B_2$ for $X$ such that $\mathcal B_1$ is a basis for $M$.

However, I wonder whether or not we can find such a basis $\mathcal B$ in a way that the projection operator $P:X \to M$ defined by $$ P(u_1+u_2)=u_1,\ \mbox{for every } u_1\in spam(\mathcal B_1),\ u_2\in spam(\mathcal B_2), $$ satisfies $\|P\|\leq 1$.

André Porto
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1 Answers1

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Very nice question.

As far as I can see, the equivalence between (i) and (iii) in Theorem 3.1 of this survey paper (https://projecteuclid.org/download/pdf_1/euclid.twjm/1500574888) shows that if your space has dimension at least 3, then your claim holds if and only if it is isometrically isomorphic to a Hilbert space, in which case the claim is trivial.

PhoemueX
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  • I really appreciated your reference. The result you pointed out is almost close enough to answer my question. However, I'm only requiring the existence of a projection with norm at most $1$, not exactly $1$. Anyway, this is pretty good survey. – André Porto Dec 05 '19 at 15:53
  • @Andre: The norm of a nonzero projection operator is always at least one, simply because $Px = x$ for any $x$ in the range of $P$, so this should be fine :) – PhoemueX Dec 05 '19 at 19:35
  • Indeed! Ok, now I follow. So the only interesting case is when $X$ has dimension $2$. When we are working with 2-dimensional spaces, it's really just the matter of picking some unit vector $x$ in $M$ and choosing $y$ such that $B_X$ is contained in the paralelogram $A={ax+by: |a|, |b|\leq 1}$. This way the projection $P(ax+by)=ax$ will have norm 1, because $P(B_X)\subset P(A)\subset{\alpha x: \alpha\in[-1,1]}$. – André Porto Dec 05 '19 at 21:03