Show that P is an L-projection iff $P^{*}$ is an M-projection

Question

I have started reading "M-ideals in Banach spaces and Banach algebras", but I stuck on the first page. It says that "there is an obvious duality between L- and M- projections:

• P is an L-projection on X iff $P^{*}$ is an M-projection on $X^{*}$
• P is an M-projection on X iff $P^{*}$ is an L-projection on $X^{*}$" ,

that is (as for the 1st sentence) :

$$ \forall_{x\in X}\Vert x\Vert=\Vert Px\Vert+\Vert x-Px\Vert\implies \forall_{f\in X^{*}} \Vert f\Vert=max(\Vert P^*f\Vert,\Vert f-P^*f\Vert) $$ I have tried but it is not so obvious for me...

Answer 1

Let $P$ be an $L$-projection. From

$$\lVert x\rVert = \lVert Px\rVert + \lVert x-Px\rVert$$

for all $x\in X$ it follows that $\lVert P\rVert \leqslant 1$ and $\lVert I-P\rVert \leqslant 1$, and hence

$$\lVert P^\ast f\rVert = \lVert f\circ P\rVert \leqslant \lVert f\rVert\cdot \lVert P\rVert \leqslant \lVert f\rVert,$$

and analogously $\lVert (I-P)^\ast f\rVert \leqslant \lVert f\rVert$ for all $f\in X^\ast$. Therefore $\lVert f\rVert \geqslant \max \left\{ \lVert P^\ast f\rVert, \lVert f - P^\ast f\rVert\right\}$. On the other hand,

$$\begin{align} \lvert f(x)\rvert &= \lvert f(Px) + f(x-Px)\rvert\\ &\leqslant \lvert f(Px)\rvert + \lvert f((I-P)x)\rvert\\ &= \lvert f(P^2x)\rvert + \lvert f((I-P)^2x)\rvert\\ &= \lvert (P^\ast f)(Px)\rvert + \lvert ((I-P)^\ast f)((I-P)x)\rvert\\ &\leqslant \lVert P^\ast f\rVert\lVert Px\rVert + \lVert f-P^\ast f\rVert\lVert x-Px\rVert\\ &\leqslant \max\left\{ \lVert P^\ast f\rVert, \lVert f-P^\ast f\rVert\right\} \left(\lVert Px\rVert + \lVert x-Px\rVert\right)\\ &= \max\left\{ \lVert P^\ast f\rVert, \lVert f-P^\ast f\rVert\right\} \lVert x\rVert, \end{align}$$

whence $\lVert f\rVert \leqslant \max\left\{ \lVert P^\ast f\rVert, \lVert f-P^\ast f\rVert\right\}$.

Together, we have $\lVert f\rVert = \max\left\{ \lVert P^\ast f\rVert, \lVert f-P^\ast f\rVert\right\}$, so $P^\ast$ is an $M$-projection.

Let now $P$ be an $M$-projection. Since $f = P^\ast f + (f-P^\ast f)$, it follows that $\lVert f\rVert \leqslant \lVert P^\ast f\rVert + \lVert f-P^\ast f\rVert$ for all $f\in X^\ast$. For the reverse inequality, let $0 < \varepsilon < 1$, and choose $a,b\in X$ with $\lVert a\rVert\leqslant 1$, $\lVert b\rVert\leqslant 1$ and

$$(P^\ast f)(a) \geqslant (1-\varepsilon)\lVert P^\ast f\rVert,\quad ((I-P)^\ast f)(b) \geqslant (1-\varepsilon)\lVert f-P^\ast f\rVert.$$

Let $x = Pa + (I-P)b$. Then

$$\lVert x\rVert = \max \left\{ \lVert Px\rVert, \lVert x-Px\rVert\right\} = \max \left\{ \lVert Pa\rVert, \lVert (I-P)b\rVert\right\} \leqslant \max \{ \lVert a\rVert, \lvert b\rVert\} \leqslant 1,$$

and

$$\begin{align} f(x) &= f(Pa + (I-P)b)\\ &= f(Pa) + f((I-P)b)\\ &= (P^\ast f)(a) + ((I-P)^\ast f)(b)\\ &\geqslant (1-\varepsilon)\left(\lVert P^\ast f\rVert + \lVert f-P^\ast f\rVert\right), \end{align}$$

hence $\lVert f\rVert \geqslant (1-\varepsilon)\left(\lVert P^\ast f\rVert + \lVert f-P^\ast f\rVert\right)$. Since $\varepsilon$ was arbitrary, it follows that $\lVert f\rVert \geqslant \lVert P^\ast f\rVert + \lVert f-P^\ast f\rVert$. Together, we have equality, and $P^\ast$ is seen to be an $L$-projection.

For the implications in the other direction, note that if $P^\ast$ is an $L$- resp. $M$-projection, then by the above $P^{\ast\ast}$ is an $M$- resp. an $L$-projection, and since $P$ is the restriction of $P^{\ast\ast}$ to $X$, it follows that $P$ is an $M$- resp. an $L$-projection.

---

From a more systematic viewpoint, if $R$ and $K$ are the range and kernel of $P$ respectively, then $X$ is isomorphic to the direct product $R\times K$ via the map $\Phi \colon x \mapsto (Px, x-Px)$. Then

• $P$ is an $L$-projection if and only if $\Phi$ is an isometry when the product $R\times K$ is endowed with the sum norm, $\lVert (u,v)\rVert_1 = \lVert u\rVert + \lVert v\rVert$, and
• $P$ is an $M$-projection if and only if $\Phi$ is an isometry when the product $R\times K$ is endowed with the maximum norm, $\lVert (u,v)\rVert_\infty = \max \left\{ \lVert u\rVert, \lVert v\rVert\right\}$.

The correspondence between $L/M$-projections on $X$ and $M/L$-projections on $X^\ast$ then is a manifestation of the fact that the dual norm of the sum norm on $R\times K$ is the maximum norm on $R^\ast\times K^\ast$ and the dual norm of the maximum norm is the sum norm on $R^\ast \times K^\ast$.