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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.

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  • $\begingroup$ 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 $(X,\lVert\cdot\rVert_1)$ $\endgroup$
    – Lorago
    Mar 22, 2023 at 15:11
  • $\begingroup$ @Lorago Yes, $M$ being closed will make that happen. I am wondering if that condition is actually needed. $\endgroup$
    – ashK
    Mar 22, 2023 at 15:12

1 Answer 1

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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 $n_0\in N.$

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