I want to prove the following theorem.
Theorem Let $X_1$ be a closed subspace of a Banach space $X$ and $M$ be a closed with $X$ such that $M\cap X_1$={0}. Then $X_2=X_1\oplus M$ is a closed subspace of $X$. Moreover, the operator $P$ defined by \begin{align*} Px=\left\{ \begin{array}{cc} x&(x\in M)\\ 0&(x\in X_1) \end{array} \right. \end{align*} then, $P\in B(X)$. ($B(X)$ is the set of bounded linear functionals from $X$ to $X$ whose domain of definition is the whole $X$.)
My view
Suppose it was shown that $P$ is a bounded operator on $X_2$. Then we show that $X_2$ is closed. Assuming $x_n\in X_2$ and $x_n\to x$ in $X$, from the boundedness of $P$, $\{Px_n\}_{n=1}^{\infty}$ is a Cauchy sequence of $M$. Since $M$ is a closed subset of Banach space, it is complete, so $\{Px_n\}_{n=1}^{\infty}$ converges to some $m\in M$. Therefore, $\{(I-P)x_n\}_{n=1}^{\infty}$ converges to $x-m$. Call this $z$: $z=x-m\in X_1$. Thus, $x_n=Px_n+(I-P)x_n\to m+z\in X_2$, and since $x=m+z$, $x\in X_2$, and $X_2$ is closed.
The problem here is the boundedness of $P$. From the definition of the operator norm, I want to compute $\|P\|$, but I can't show that it is finite. \begin{align*} \|P\|&=\sup_{x\in M_2,x\neq 0}\frac{\|Px\|}{\|x\|}=\sup_{m+x_1\neq 0,m\in M,x_1\in X_1}\frac{\|m\|}{\|m+x_1\|}=\sup_{\|m\|=1,m\in M,x_1\in X_1}\frac{1}{\|m+x_1\|} \end{align*} I think we can transform this far, but we need to show that no matter how we take $m\in M,x_1\in X_1$, this denominator will not be $0$.
