Here's my question:
Suppose $(X , \lVert \cdot \rVert)$ is a Banach Space, $M$ be a
subspace of $X$. Assume that $N$ is closed subspace of $X$ such that
- $M + N = X$ and
- $M \cap N = \{ 0 \}$ Then consider the space $(X, \lVert \cdot \rVert_{1})$ given by $$\begin{align*} \lVert m+n \rVert _1 = \lVert m
\rVert + \lVert n \rVert \end{align*}$$ for each $m\in M$ and for each
$n \in N$.Is $(X , \lVert \cdot \rVert_1 )$ a Banach Space?
I suspect that it is. However I am unable to complete its proof. Here's my incomplete attempt:
It is easy to see that $(X, \lVert \cdot \rVert)_1$ is a normed linear space. We proceed to show that it is a Banach space.
- Let $(x_n)$ be a Cauchy sequence in $(X, \lVert \cdot \rVert_1)$. By the unique decomposition, we have that for each $k \in \mathbb N$, $x_k = m_k + n_k$ for some $m_k \in M$ and $n_k \in N$.
- Then we have for $k,j \in \mathbb N$
$$\begin{align*}\lVert x_k – x_j \rVert_{1} = \lVert m_k – m_j \rVert + \lVert n_k – n_j \rVert \end{align*}$$ - This shows that $(m_k)$ and $(n_k)$ are Cauchy in $X$.
- Since $N$ is closed subspace of $X$, so $(n_k)$ converges to some $n \in N$ and since $X$ is a Banach space, $(m_k)$ converges to $m \in X$.
- Also, since $\lVert y \rVert _1 \ge \lVert y \rVert$ for each $y \in X$, $(x_n)$ is Cauchy in $(X, \lVert \cdot \rVert )$. Since $(X, \lVert \cdot \rVert )$ is a Banach Space, we have that $(x_n)$ converges to $x \in X$ in $\lVert \cdot \rVert$-norm.
- Let $x=\mu + \nu$ for some $\mu \in M$ and some $\nu \in M$.
If I could show that $m = \mu$ and $n=\nu$ holds then I would be done. Hints are appreciated.
Best Answer
If $(X,\|\cdot\|_1)$ is a Banach space then $M$ is closed. Indeed, we have for $x=m+n$ $$\|x\|_1=\|m\|+\|n\|\ge \|m+n\|=\|x\|$$ Therefore the mapping $T:(X,\|\cdot\|_1)\to (X,\|\cdot\|)$ given by $Tx=x$ is a bounded bijection. By the Banach inverse mapping theorem, the inverse mapping is bounded. Hence there exists a constant $c>0,$ such that for $x=m+n$ there holds $$\|m\|+\|n\|=\|x\|_1\le c\|x\|=c\|m+n\|\quad (*)$$ Assume $m_k\to v$ for $m_k\in M.$ Let $v=m_0+n_0.$ Then $(*)$ implies $$\|m_k-m_0\|+\|n_0\| \le c\|(m_k-m_0)-n_0\|=c\|m_k-v\|\to 0$$ Hence $n_0=0$ and $m_k\to m_0,$ which shows that $M$ is closed. Observe that we haven't used the fact that $N$ is closed. Clearly the above reasoning implies that $N$ must be closed.
The converse implication holds as well. Indeed, assume $M$ and $N$ are closed and $X=M+N.$ Then $(X,\|x\|_1)$ is a Banach space, because if $x_k=m_k+n_k$ is a Cauchy sequence with respect to $\|\cdot\|_1$ norm then $m_k$ and $n_k$ are Cauchy sequences with respect to $\|\cdot\|.$ Hence they are convergent to $m_0$ and $n_0,$ respectively. Since $M$ and $N$ are closed, we get $m_0\in M$ and $n_0\in N.$