"Every linear mapping on a finite dimensional space is continuous"

Solution 1:

The result we can show is the following:

Let $E$ and $F$ two topological vector spaces, and $T\colon E\to F$ a linear map. If $E$ is finite dimensional, then $T$ is continuous.

First, if $(e_1,\ldots,e_n)$ is a basis of $E$, then any set of $n+1$ vectors of $T(E)$ is linearly dependent, so $T(E)$ has a dimension $\leqslant n$. Let $k$ be the dimension of $T(E)$, and $(v_1,\ldots,v_k)$ a basis of this space. We can write for any $x\in E$: $T(x)=\sum_{i=1}^ka_i(x)v_i$ and since $v_i$ is a basis each $a_i$ is linear. We have to show that each map $T_i\colon E\to F$, $T_i(x)=:a_i(x)v_i$ is continuous.

Added: the map $x\mapsto a_i(x)$ is well-defined because $(v_1,\ldots,v_k)$ is a basis. In particular, it takes finite values.

By definition of a topology on a topological vector space we only have to show that the map $x\mapsto a_i(x)$ is continuous. To do that, we use the fact that a finite dimensional topological vector space can be equipped with a norm which gives the same topology (in fact it's the unique one), namely put $$N\left(\sum_{j=1}^n\alpha_jx_j\right):=\sum_{j=1}^n|\alpha_j|.$$ Now the continuity is easy to check: denoting $x=\sum_{j=1}^nx_je_j$ and $y=\sum_{j=1}^ny_je_j$ $$|a_i(x)-a_i(y)|\leqslant \sum_{j=1}^n|a_i((x_j-y_j)e_j)|=\sum_{j=1}^n|x_j-y_j|\cdot |a_i(e_j)|\leqslant N(x-y)\sum_{j=1}^n|a_i(e_j),$$ since $|x_j-y_j|\leqslant N(x-y)$ for all $1\leqslant j\leqslant n$.

Solution 2:

There is a nice write-up of a proof of a simpler version of this (for normed vector spaces) on Wikipedia. Let me reproduce the proof given on Wikipedia:

If $T:X\to Y$ is a linear map where $X,Y$ are normed vector spaces and $X$ has finite dimension then $T$ is continuous.

Proof: Let $e_1,\dots, e_n$ be a basis of $X$. Then for $x \in X$ we have

$$ \|Tx\|_Y = \|T \sum_i \alpha_i e_i\|_Y = \|\sum_i \alpha_i Te_i \|_Y \le \sum_i |\alpha_i|\|Te_i\|_Y$$

Let $\varepsilon >0$ be given and let $M = \max_i \|Te_i\|_Y$. Let $\delta = {\varepsilon \over M}$. Then for $x$ with $\|x\| < \delta$ we have

$$ \|Tx\|_Y \le \sum_i |\alpha_i|\|Te_i\|_Y \le M \sum_i |\alpha_i| < \varepsilon$$

hence $T$ is continuous.

Solution 3:

The special case of a linear transformations $A: \mathbb{R}^n \to \mathbb{R}^n$ being continuous leads nicely into the definition and existence of the operator norm of a matrix as proved in these notes.

To summarise that argument, if we identify $M_n(\mathbb{R})$ with $\mathbb{R^{n^2}}$, and suppose that $v \in \mathbb{R}^n$ has co-ordinates $v_j$, then by properties of the Euclidean and sup norm on $\mathbb{R}^n$ we have:

$\begin{align}||Av|| &\leq \sqrt{n} \,||Av||_{\sup} \\&= \sqrt{n}\max_i\bigg|\sum_{j}a_{ij}\,v_j\bigg|\\&\leq \sqrt{n}\max_i \sum_{j}|a_{ij}\,v_j|\\&\leq \sqrt{n} \max_i n\big(\max_j|a_{ij} v_j|\big)\\&=n\sqrt{n} \max_i \big(\max_j |a_{ij}| \max_j |v_j|\big)\\&= n\sqrt{n}\max_{i,j}|a_{ij}|||v||_{\sup}\\&\leq n \sqrt{n} \max_{i,j}|a_{ij}|||v|| \end{align}$

$\Rightarrow ||Av|| \leq C ||v||$ where $C = n\sqrt{n}\displaystyle\max_{i,j}|a_{ij}|$ is independent of $v$

So if $\varepsilon>0$ is given, choose $\delta = \dfrac{\varepsilon}{C}$ and for $v, w \in \mathbb{R}^n$ with $||v-w||< \delta$ consider

$||Av - Aw || = ||A(v-w) || \leq C ||v-w || < \delta C= \varepsilon$ from which we conclude that $A$ is uniformly continuous.