Assume that we have finitely many projections $P_1,\dots, P_n$ on a Banach space $X$ (take an explicit case of $X=L_p(\mu)$ if a concrete examples is better). Consider their convex combination $P=\sum_{i=1}^n \alpha_i P_i$, where $\alpha_i$ are strictly positive and sum up to 1.

Question: When is the kernel of $P$ is equal to the intersection of the kernel of the $P_i$?

The only case I know how to deal with so far is when the $P_i$ have norm one, but I'd like to know how to remove this assumption. Any help will be appreciated!

  • 1
    $\begingroup$ This is incorrect. You can find two projections in the plane without this property. $\endgroup$
    – Uri Bader
    Sep 16, 2016 at 13:20
  • $\begingroup$ Agreed, in general this does not hold, but what I mean here is are there conditions that could make this hold and why - I'll edit the question to make this clear $\endgroup$
    – user98548
    Sep 16, 2016 at 13:31
  • $\begingroup$ I think that you need to explain clearly: (1) What spaces are you talking about? (I ask this because you are talking about norm 1.) (2) In what terms do you hope/want to get an answer? $\endgroup$ Sep 16, 2016 at 23:00
  • $\begingroup$ Good point, I edited the question. As for what terms I want the answer in I really can't say much. At this point I am stuck on this and any kind of condition implying this would be interesting. $\endgroup$
    – user98548
    Sep 17, 2016 at 3:00

1 Answer 1


This answer is almost trivial, but I decided to write it.

First sufficient condition for $Ker(P)=Ker(P_1)\cap...\cap Ker(P_n)$: the subspaces $Ran(P_1),...,Ran(P_n)$ are linearly independent (i.e., if $y_1+...+y_n=0$, where $y_1\in Ran(P_1)$,...,$y_n\in Ran(P_n)$, then $y_1=...=y_n=0$). The proof is trivial.

Second sufficient condition for $Ker(P)=Ker(P_1)\cap...\cap Ker(P_n)$: there exists a positive number $r$ such that the following conditions:

  1. $\|rI-P_i\|\leqslant r$;

  2. if $\|(rI-P_i)x\|=r\|x\|$, then $(rI-P_i)x=rx$ (i.e., $P_i x=0$)

hold for $i=1,...,n$.

Indeed, suppose that $Px=0$. Then $\alpha_1 P_1 x+...+\alpha_n P_n x=0$ and, consequently, $\alpha_1(rI-P_1)x+...+\alpha_n(rI-P_n)x=rx$. But \begin{align*} &\|\alpha_1(rI-P_1)x+...+\alpha_n(rI-P_n)x\|\leqslant\\ &\alpha_1\|(rI-P_1)x\|+...+\alpha_n\|(rI-P_n)x\|\leqslant\\ &\alpha_1 r\|x\|+...+\alpha_n r\|x\|=r\|x\|. \end{align*} Hence, $\|(rI-P_i)x\|=r\|x\|$ for all $i=1,...,n$. Consequently, $P_i x=0$, $i=1,...,n$.


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