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Let $c$ be the set of all convergent real sequences $\mathbb N \to \mathbb R$ and $c_0$ be the subspace of all the sequences that converges to zero. We can consider $c$ as a Banach space under the norm $||\cdot||_{\infty}$, $c_0$ it's also a Banach space because it's a closed subspace of the complete space $c$.

I want to find a projection $P: c\to c_0$ (i.e linear, continuous and such that $P^2=P$) and prove that this projection has norm always greater than 1.

We are under the assumption that $c_0\subset c \subset l^{\infty}$ ($l^{\infty}$ denotes the bounded sequences endowed with the $||\cdot||_{\infty}$ norm.

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  • $\begingroup$ You need your sequences to be at least bounded for $\|\cdot\|_\infty$ to be defined. $\endgroup$ – Martin Argerami Oct 15 '13 at 0:29
  • $\begingroup$ I think you want $P$ to be surjective. (That is what I assumed in my answer.) $\endgroup$ – Jonas Meyer Oct 16 '13 at 0:57
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$c$ is the space of convergent sequences (it said this in the title, and I just edited your question to include it). Thus it makes sense to define $P:c\to c_0$ by $P(x_0,x_1,x_2,\ldots)=(x_0-\lim\limits_{n\to\infty}x_n,x_1-\lim\limits_{n\to\infty}x_n,x_2-\lim\limits_{n\to\infty}x_n,\ldots)$.

Regardless of choice, if $P^2=P$ and $P(c)=c_0$, then there exists $x\in c$ such that $Px\neq 0$, and $P(Px)=Px$ implies that $\|P\|\geq 1$.

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  • $\begingroup$ (In the example $\|P\|=2$.) $\endgroup$ – Jonas Meyer Oct 15 '13 at 18:39
  • $\begingroup$ How can I prove that there exist $x$ such that $Px\ne 0$? $\endgroup$ – Shanks Oct 15 '13 at 23:18
  • $\begingroup$ @Shanks: $Px=0$ for all $x$ would mean that $P=0$, by definition. I realize that you never stated it explicitly, but I assumed you wanted $P$ to be surjective, i.e., $P(c)=c_0$, or equivalently $Px=x$ for all $x\in c_0$, not to mention that you want $P\neq 0$. (Of course if $P$ were $0$ you would have $\|P\|=0$.) Because $c_0\neq\{0\}$, any $x\in c_0\setminus\{0\}$ will do. $\endgroup$ – Jonas Meyer Oct 16 '13 at 1:00
  • $\begingroup$ @JonasMeyer for a surjective projection $P$, shouldn't $\| P\| \ge 2$? since such a $P$ must map $Px = x$ for any $x \in c_0$? $\endgroup$ – dimebucker Sep 9 '18 at 1:27
  • $\begingroup$ @dimebucker: No. $Px = x$ for $x\neq 0$ only implies that $\|P\| \geq 1$, from the definition. The identity operator has norm $1$ on every normed space, for example, as do all self-adjoint projections on Hilbert space. $\endgroup$ – Jonas Meyer Oct 2 '18 at 19:42

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