= Completeness of a graph norm from a bounded right inverse
{title2=$\ker T\text{ complete},\ TS=I,\ S\text{ bounded}\ \Longrightarrow\ (X,\|\cdot\|_T)\text{ complete}$}
Let $X$ be a <normed space>, $Y$ a <Banach space>, and $T:X\to Y$ a <linear map>. If its <kernel> is <complete> in the original <norm> and $S:Y\to X$ is a bounded <right inverse>, then the <graph norm> $\|x\|_T=\|x\|_X+\|Tx\|_Y$ is <complete>, even if $X$ with its original <norm> is incomplete or $T$ is unbounded.
For a <Cauchy sequence> $(x_n)$ in the <graph norm>, $y_n=Tx_n$ converges to $y\in Y$. The elements $w_n=x_n-Sy_n$ lie in the <kernel> and satisfy $\|w_n-w_m\|_X\leq(1+\|S\|)\|x_n-x_m\|_T$. Hence $w_n\to w\in\ker T$. The element $x=w+Sy$ satisfies $Tx=y$ and
$$
\|x_n-x\|_T\leq\|w_n-w\|_X+(1+\|S\|)\|y_n-y\|_Y\longrightarrow0.
$$
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