To verify whether Banach space

Question

($X,||.||$) is a Banach space and $X=M*N$, where M and N are closed subspaces of X. Let $x \in X$ have a unique representation $x=m+n$, $m \in M$ and $n \in N$ Define $||.||_{1}$ : $X \to R$ by $||x||_{1}$ = $||m||$ + $||n||$, $x \in X$

We need to verify that $(X, ||.||_{1})$ is a Banach space.

This is what I have done so for. Since, $M$ and $N$ are closed subspaces of $X$ and $X$ is Banach so $M$ and $N$ are Banach space $w.r.t$ $||.||$

Now, $||x||_{1} = ||m|| + ||n|| >=0$ and $||x||_{1} = 0$ $iff$ $||m||=0$ and $||n||=0$ $iff$ $m=0$ and $n=0$ $i.e.,$ $x=m+n=0$

Now,$||cx||_{1} = ||cm|| + ||cn|| = |c|||m|| + |c|||n|| = |c|(||m||+||n||)=|c|||x||_{1}$

And, $||x+y||_{1}$ = $||m_{1} + n_{1} + m_{2} +n_{2}||$

But from here I could not proceed any further. I know I need to show $(X,||.||_{1})$ is a normed linear space and show that X is complete. Can you please help me with how to proceed further?

Answer 1

If $(x_k\equiv m_k+n_k)$ is a Cauchy sequence then $\|x_k-x_j\|_1=\|m_k+n_k-(m_j+n_j)\|_1=\|m_k-m_j\|+\|n_k-n_j\|\to 0$. This implies that $(m_k)$ and $(n_k)$ are Cauchy sequences, so they converge to some points $m \in M$ and $n \in N$ (because $M$ and $N$ are closed). Can you finish?