Projection in Banach Space

Question

Let $X$ be a Banach Space and let $Y=\ker f \subset X$ be hyperplane in $X$. Prove that there exists a projection $P:X \to Y$ such that $||P||\leq 2$.

Answer 1

Let $g=\Vert f\Vert^{-1} f$. Then $\ker(g)=\ker(f)$. Using one of the corollaries of the Hahn-Banach theorem choose $x_0\notin \ker(g)$ such that $g(x_0)=1$ and $\Vert x_0\Vert=1$. Then define $$ P:X\to X:x\mapsto x-g(x)x_0 $$ Let's check that this is a well defined projection onto $Y$ with norm not greater than 2. For all $x\in X$ we have $$ \begin{align} P^2(x)&=P(P(x))\\ &=P(x-g(x)x_0)\\ &=P(x)-P(g(x)x_0)\\ &=P(x)-g(x)P(x_0)\\ &=P(x)-g(x)(x_0-g(x_0)x_0)\\ &=P(x)-g(x)(x_0-x_0)\\ &=P(x) \end{align} $$ Therefore $P$ is a projection.

For any $x\in X$ we have $x=y+tx_0$ for some $y\in Y$, $t\in\mathbb{K}$. Since $y\in Y=\ker(g)$, then $$ P(x)=y+tx_0-g(y+tx_0)x_0=y+tx_0-g(y)x_0-tg(x_0)x_0=y+tx_0-tx_0=y\in Y $$ Therefore $P$ is a projection onto $Y$. And the last note $$ \begin{align} \Vert P(x)\Vert&=\Vert x-g(x)x_0\Vert\\ &\leq\Vert x\Vert+\Vert g(x)x_0\Vert\\ &=\Vert x\Vert+|g(x)|\Vert x_0\Vert\\ &\leq\Vert x\Vert+\Vert g\Vert\Vert x\Vert\Vert x_0\Vert\\ &\leq\Vert x\Vert+\Vert x\Vert\\ &=2\Vert x\Vert \end{align} $$ Therefore $\Vert P\Vert\leq 2$