Problem 5

Exercise 8.5 (Projection operator and closed subspaces)   Let $X=X_1\dot{+}X_2$ and define $P(x_1+x_2)=x_1$. Then $P$ is a linear operator satisfying $P^2=P$ and moreover, $P$ is bounded if and only if both $X_1$ and $X_2$ are closed.

Proof. To see $P$ is a linear operator, let $x,y\in X$ decompose uniquely into $x=x_1+x_2$ and $y=y_1+y_2$. Then $x+y=(x_1+y_1) + (x_2+y_2)$ uniquely, so we see

$$P(x+y) = x_1+y_1 = Px_1 + Py_1 = Px + Py$$ (8.24)

Moreover, $P^2=P$ because if $x\in X$, then $x=x_1+x_2$ uniquely. Projecting, we know $Px=x_1+0$ uniquely, so that $P(Px)=x_1=Px$. Therefore, $P^2=P$.

Now to see $P$ is bounded implies $X_1$ and $X_2$ are closed, we just write them as follows

$$X_1 = P(X)$$ (8.25)

$$X_2 = \ker P$$ (8.26)

The closed graph theorem applies, so that

$$\Gamma = \{(x_1+x_2,x_1) \;\vert\; x_1\in X_1, x_2\in X_2\} = X\times X_1$$ (8.27)

is a closed subspace of $X\times X$, indicating $X_1$ is a closed subspace. Kernels of bounded operators are closed, so $X_2$ is closed.

Conversely, if $X_1$ and $X_2$ are closed subspaces, they are themselves Banach spaces, so we may define the direct sum of $Y=X_1\oplus X_2$ under the norm

$$\Vert(x_1,x_2)\Vert=\Vert x_1\Vert _{X_1}+\Vert x_2\Vert _{X_2}$$ (8.28)

Completeness is inherited from the completeness of $X_1$ and $X_2$. For any $x=x_1+x_2\in X$, we know $\Vert x\Vert _X\leq \Vert(x_1,x_2)\Vert _Y$, so this shows $X\cong Y$ by mapping $x_1+x_2\mapsto(x_1,x_2)$.

Note that $P$ acts on the space $Y$ by $P(x_1,x_2)=(x_1,0)$, and this means

$$\Vert P\Vert _{Y\to Y}=\sup_{y\in Y}\frac{\Vert Py\Vert}{\Vert y\... ...frac{\Vert x_1\Vert _{X_1}}{\Vert x_1\Vert _{X_1}+\Vert x_2\Vert _{X_2}} \leq 1$$ (8.29)

so $P$ is bounded as an operator on $Y$. By the isomorphism, we know $P$ is bounded as an operator on $X$. $\qedsymbol$