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Let $X$ and $Y$ be normed linear spaces and let $T:X\rightarrow Y$ be open, linear and continuous. If $X$ is complete then show that $Y$ is complete.

I have proceeded little bit. Since $T$ is an open map then $T$ is onto and there exists a $\delta >0$ such that for each $y\in Y$ there exists $x\in X$ with $\|x\|<\delta \|y\|$ and $T(x)=y$. Now let us take a sequence $(y_{n})$ in $Y$ which is Cauchy.

Now we will get $(x_{n})$ such that $\|x_{n}\|<\delta \|y_{n}\|$ and $T(x_{n})=y_{n}$. Now if we can show that $x_{n}$ is Cauchy, the rest will follow. But I can't show that.

Riju
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Generally, you won't have $\{x_n\}$ convergent; but you don't need them to be. The goal should be to get them to converge modulo $\ker T$; that is, in the quotient space $X/\ker T$. Let's write $M=\ker T$ for brevity; this is a closed subspace of $X$. The norm of a coset of $M$ is defined as $\|x+M\|=\inf_{y\in M}\|x+y\| = \operatorname{dist}(x,M)$.

The main step is to prove that $X/M$ is complete. This is easily done using the characterization of Banach spaces as the normed spaces where absolute convergence of a series implies convergence. Indeed, if $\sum_{n} \|x_n+M\|<\infty$, then pick a representative $x_n'\in x_n+M$ such that $\|x_n' \|<\|x_n+M\| + 2^{-n}$ and conclude that $\sum x_n' = s$ converges by the completeness of $X$; hence $\sum (x_n+M) = s+M$.

The openness of $T$ implies $\|x+M\| \le \delta \|Tx\|$ for all $x\in X$. Given a Cauchy sequence $\{y_n\}$, pick any $x_n$ such that $Tx_n=y_n$. Then the sequence $\{x_n+M\}$ is Cauchy in $X/M$, hence it has a limit $x+M$. It follows that $y_n=Tx_n\to Tx$ as required.

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  • You can also see that $X/M$ is complete in another way.

    Lemma: Let $M$ be closed in $X$. $, X$ is Banach iff $M$ is Banach and $X/M$ is Banach. $ \ $ Here, $M$ is closed as $T$ is continuous and $M$ is the inverse image of $ 0 \in \mathbb{R} $

     – Adithya S Dec 07 '20 at 17:02