linear transformation is continuous if its domain is finite dimensional

Theorem 1.

A linear transformation is continuous if the domain is finite dimensional.

Proof.

Suppose $L:X\to Y$ is the transformation, $dimX=n$, and $\parallel \cdot {\parallel }_X$, $\parallel \cdot {\parallel }_Y$ are the norms on $X$, $Y$, respectively. By this result (http://planetmath.org/ContinuityIsPreservedWhenCodomainIsExtended) and this result (http://planetmath.org/SubspaceTopologyInAMetricSpace), it suffices to prove that $L:X\to L(X)$ is continuous when $L(X)$ is equipped with the topology given by $\parallel \cdot {\parallel }_Y$ restricted onto $L(X)$. Also, since continuity and boundedness are equivalent, it suffices to prove that $L$ is bounded. Let $e_1,\mathrm{\dots },e_n$ be a basis for $X$ such that $L$ is invertible on $\mathrm{span}\{e_1,\mathrm{\dots },e_k\}$ and $\ker L=\mathrm{span}\{e_{k+1},\mathrm{\dots },e_n\}$ for $k=1,\mathrm{\dots },n$. (The zero map is always continuous.) Let $f_i=L(e_i)$ for $i=1,\mathrm{\dots },k$, so that $\mathrm{span}\{f_1,\mathrm{\dots },f_k\}=L(X)$. Let us define new norms on $X$ and $L(X)$,

${\parallel x\parallel }_X^{\prime }$ $=$ $\sqrt{{\displaystyle \sum _{i=1}^n}{\alpha }_i^2},$

${\parallel y\parallel }_X^{\prime }$ $=$ $\sqrt{{\displaystyle \sum _{i=1}^k}{\beta }_i^2},$

for $x={\sum }_{i=1}^n{\alpha }_i{e}_i\in X$ and $y={\sum }_{i=1}^k{\beta }_i{f}_i\in Y$. Since norms on finite dimensional vector spaces are equivalent, it follows that

$1/C{\parallel x\parallel }_X^{\prime }\le {\parallel x\parallel }_X\le C{\parallel x\parallel }_X^{\prime },x\in X$

$1/D{\parallel y\parallel }_Y^{\prime }\le {\parallel y\parallel }_Y\le D{\parallel y\parallel }_Y^{\prime },y\in L(X)$

for some constants $C,D>0$. For $x={\sum }_{i=1}^n{\alpha }_i{e}_i\in X$,

${\parallel L(x)\parallel }_Y$ $\le$ $D{\parallel {\displaystyle \sum _{i=1}^k}{\alpha }_i{f}_i\parallel }_Y^{\prime }$

$=$ $D\sqrt{{\displaystyle \sum _{i=1}^k}{\alpha }_i^2}$

$\le$ $D\sqrt{{\displaystyle \sum _{i=1}^n}{\alpha }_i^2}$

$=$ $D{\parallel x\parallel }_X^{\prime }$

$=$ $CD{\parallel x\parallel }_X.$

Thus $L:X\to L(X)$ is bounded. ∎

Title	linear transformation is continuous if its domain is finite dimensional
Canonical name	LinearTransformationIsContinuousIfItsDomainIsFiniteDimensional
Date of creation	2013-03-22 15:17:59
Last modified on	2013-03-22 15:17:59
Owner	matte (1858)
Last modified by	matte (1858)
Numerical id	7
Author	matte (1858)
Entry type	Theorem
Classification	msc 15A04