# Proving or disproving a space is a Banach Space

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.

• If you can show that $m\in M$, then $\lVert x_j-(m+n)\rVert_1=\lVert m_j-m\rVert+\lVert n_j-n\rVert\to0$ as $j\to\infty$, and you can conclude that $x_j\to m+n$ in $(X,\lVert\cdot\rVert_1)$ Mar 22, 2023 at 15:11
• @Lorago Yes, $M$ being closed will make that happen. I am wondering if that condition is actually needed.
– ashK
Mar 22, 2023 at 15:12

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 $$n_0\in N.$$