Why the norm is $>1$ for one projection operator? Let $P: c \to c$ be a projection operator onto $c_0 \subseteq c$. That is, $P$ is a bounded operator such that $P^2=P$ and $\operatorname{Im} P=c_0$. We are asked to show that $\lVert P\rVert > 1$.
My attempt was:
For any $w\in c$ we have that
$$\lVert Pw \rVert= \lVert P^2 w\rVert \leq \lVert P\rVert^2 \lVert w\rVert$$ this implies that $\frac{\lVert Pw \rVert}{\lVert w\rVert}\leq \lVert P\rVert^2$ or equivalently $1\leq \lVert P\rVert$. I do know why the inequality is strict. Hence, the other side we know that for all $w\in c $ then $w=w-P(w)+P(w)$ where $w-P(w)\in \ker P$ and $P(w)\in c_0= \operatorname{Im} P$, so $\lvert c \rvert = \lvert \ker P \rvert + \lvert c_0 \rvert > \lvert c_0\rvert$ where i mean $\lvert c \rvert := \operatorname{dim}c$. I guess that I need to find some particular element in the domain for attach the grater dimension or something like that. I am not sure if $c_0$ are the sequence that converge to $0$ for this reason i did no try other thing with sequences. I will appreciate any hint please.
Best
 A: Edit: we always have $\|P\|\geq2$. See Ryszard's great answer.
(my older answer below)

Let $y=P1$.  We consider two cases:

*

*If $\|y\|>1$, then $\|P1\|=\|y\|>1=\|1\|$, and we get $\|P\|>1$.


*If $\|y\|\leq1$, fix $k$ such that $|1-y_k|>\frac12$ (such $k$ exists because $y\in c_0$). Let $(1-y_k)=|1-y_k|\,e^{i\theta}$ be the polar decomposition. Let
$$
x=-\frac14\,(1-y)+e^{i\theta}\,e_k. 
$$
We have $Px=e^{i\theta}\,e_k$, so $\|Px\|=1$. And
$$
\|x\|=\max\big\{\max\{\frac14|1-y_j|:\ j\ne k\},\big|-\frac14\,|1-y_k|+1\big|\,\big\}.
$$
Since $\|y\|\leq1$, we have $\frac14\,|1-y_j|\leq\frac24=\frac12$. And
$$
\big|-\frac14\,|1-y_k|+1\big|=1-\frac14\,|1-y_k|<1-\frac18.
$$
Then
$$
\|Px\|=1>1-\frac18>\|x\|,
$$
and hence $\|P\|>1$.
A: Let $P$ be a projection onto $c_0.$ We will show that $\|P\|\ge 2.$
Let $x_\infty =\lim x_n.$ Observe that $P$ is of the form $$P_vx=x-x_\infty v,\quad\ v\in c,\ v_\infty=1\quad (*)$$ Indeed. let $1{\hskip -2.5 pt}\hbox{l}$ denote the constant sequence with all terms equal $1.$ we have $$\displaylines{Px=P(x-x_\infty 1{\hskip -2.5 pt}\hbox{l})+x_\infty  P1{\hskip -2.5 pt}\hbox{l}\\ =x-x_\infty 1{\hskip -2.5 pt}\hbox{l} +x_\infty P1{\hskip -2.5 pt}\hbox{l}\\ = x-x_\infty (1{\hskip -2.5 pt}\hbox{l}-P1{\hskip -2.5 pt}\hbox{l})}$$ thus $v=1{\hskip -2.5 pt}\hbox{l}-P1{\hskip -2.5 pt}\hbox{l},$ which completes the proof of $(*).$
From now on we are dealing with $P_v$ for a sequence $v$ with $v_\infty =1.$
Let $x^{N}$ denote the sequence with terms $$x^{(N)}_n=\begin{cases} -v_n& 1\le n\le N\\ v_n & n>N\end{cases}$$ Then the limits of this sequence is equal $v_\infty,$ i.e. $x^{(N)}_\infty =v_\infty=1,$ $\|x^{(N)}\|=\|v\|.$ By $(*)$ we obtain
$$[P_v(x^{(N)})]_n = \begin{cases} -2v_n& 1\le n\le N\\ 0 & n>N\end{cases}$$ Thus $$\|P_v(x^{(N)})\|=2\max_{1\le n\le N}|v_n|\underset{N}{\rightarrow}2=2\|x^{(N)}\|$$ Hence $\|P_v\|\ge 2.$
Remark
It may occur that the norm $2$ is not attained at any element $x\in c.$ For example let $$v_n=1-{1\over n}$$ Then $$\|P_vx\|=\|x-x_\infty v\| \le 2\|x\| $$ Assume, by contradiction that there is $x\in c $ such that $$\|x-x_\infty v\|=2,\quad \|x\|=1$$ With no loss of generality we may assume $x_\infty >0.$
Then $$2=\|x-x_\infty v\|\le \|x\|+x_\infty\le 2$$ Hence $x_\infty =1$ and $$\|x-v\|=2$$ As $\lim(x_n-v_n)=0$ there is $n$ such that $|x_n-v_n|=2.$ This gives a contradiction because $|x_n|\le 1$ and $0<v_n<1.$
