Let a Banach space $B=M \oplus N$ where $M$ and $N$ are closed linear subspaces of $B$.
If $z=x+y$ is the unique representation of $z$ in $B$ where $x$ and $y$ are from $M$ and $N$ respectively then define a new norm by $\|z\|'=\|x\|+\|y\|$.
If $B'$ symbolises the linear space B equipped with this norm $\|\cdot\|'$ then show that $B'$ is a Banach space.
I have got that $M$,$N$ are complete subspaces of $B$.How do I show that a Cauchy sequence in $B'$ converges?
On
Hint: $\exists m_1,n_1\geqslant n_0$ such that $||x_{m_1}-x_{n_1}||<\frac{\epsilon}{2}\:\:\epsilon>0$, the same goes for $y\in N$ $ \exists m_2,n_2\geqslant n_2$ such that $||y_{m_2}-y_{n_2}||<\frac{\epsilon}{2}$
We can define an $\exists m,n\geqslant n_0$ such that $||x_m-x_n||<\frac{\epsilon}{2}$ and $||y_m-y_n||<\frac{\epsilon}{2}$
Since $||z_m-z_n||=||x_m-x_n||+||y_m-y_n|||$, it follows the ...
Define a norm on $M\times N$ by $$||(x,y)||=||x||+||y||.$$It's easy to see this makes $M\times N$ into a Banach space. Define $T:M\times N\to B$ by $$T((x,y))=x+y.$$Then $T$ is bounded. The fact that $B=M\oplus N$ shows that $T$ is invertible, and now the Open Mapping Theorem shows that $T^{-1}$ is bounded.
Hence $||.||'$ is equivalent to $||.||$, in the sense that $c_1||z||'\le||z||\le c_2||z||'$.