Undergrad Is the Quotient of a Banach Space by a Closed Linear Subspace also Banach?

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Here is this week's POTW:

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Prove that the quotient of a Banach space $X$ by a closed linear subspace $M$ is Banach with respect to the norm $$\|x + M\| := \inf\{\|x + y\|_X : y\in M\}\quad (x\in X)$$

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Congratulations to Opalg for his correct solution, which you can read below:

Spoiler

The wording of the question implies that the given function $\|x+M\|$ is a norm on the quotient space, and we just have to show that it makes the quotient space complete.

Let $\{x_n+M\}_{n\in\Bbb{N}}$ be a Cauchy sequence in $X/M$. There exists an increasing sequence $\{n_k\}_{k\in\Bbb{N}}$ such that $\|x_m - x_n+M\| = \|(x_m+M) - (x_n+M)\| <2^{-k}$ whenever $m,n \geqslant n_k$.

The next step is to construct a sequence $\{y_k\}$ with $y_k\in x_{n_k}+M$ such that $\|y_k - y_{k+1}\| < 2^{-k}$. Let $y_1 = x_{n_1}$. Then $$\| x_{n_1} - x_{n_2} + M\| = \inf_{z\in M} \{ x_{n_1} - (x_{n_2}+z)\| <2^{-1}.$$ If $y_2 = x_{n_2} +z$, where $z$ is sufficiently close to achieving that infimum, then $y_2\in x_{n_2} + M$ and $\|y_1-y_2\| < 2^{-1}$.

The inductive construction for the rest of the sequence is essentially the same. If $y_k \in x_{n_k}$, then $$\| x_{n_k} - x_{n_{k+1}} + M\| = \inf_{z\in M} \{ y_k - (x_{n_2}+z)\| <2^{-k},$$ and we can choose $y_{k+1} = x_{n_2}+z$ so that $\|y_k - y_{k+1}\| < 2^{-k}.$

The sequence $\{y_k\}$ is Cauchy in $X$, because if $l>k$ then $$\|y_k - y_l\| \leqslant \|y_k - y_{k+1}\| + \|y_{k+1} - y_{k+2}\| + \ldots + \|y_{l-1} - y_l\| < 2^{-k} + 2^{-(k+1)} + \ldots + 2^{-(l-1)} < \sum_{r=k}^\infty 2^{-r} = 2^{-(k-1)}.$$ Since $X$ is complete, it follows that $\{y_k\}$ converges to a limit $x$. Therefore $\|(x_{n_k}+M) -( x + M)\| \leqslant \|y_k - x\| \to0$ as $k\to\infty$. But if a subsequence of a Cauchy sequence converges, then the whole sequence converges (to the same limit). Therefore $\{x_n+M\}$ converges to $x+M$, which shows that $X/M$ is complete.