Linear Transformations between Finite-Dimensional Normed Vector Spaces are Continuous
Theorem
Linear transformations between finite-dimensional normed vector spaces are continuous.
Proof
We have that Norms on Finite-Dimensional Real Vector Space are Equivalent.
Choose the Euclidean norm.
Let $X = \struct {\R^n, \norm {\, \cdot \,}_2}$ and $Y = \struct {\R^m, \norm {\, \cdot \,}_2}$ be normed vector spaces.
Let the matrix $A \in \R^{m \times n}$ be given by:
- $A = \begin {bmatrix}
a_{1 1} & \cdots & a_{1 n} \\
\vdots & \ddots & \vdots \\
a_{m 1} & \cdots & a_{m n} \\ \end{bmatrix}$
We have that Set of Linear Transformations is Isomorphic to Matrix Space.
Let $T_A : \R^n \to \R^m$ be the linear transformation such that:
- $\forall \mathbf x \in \R^n : T_A \mathbf x := A \mathbf x$
Then:
\(\ds \norm {T_A \mathbf x}^2_2\) | \(=\) | \(\ds \norm {A \mathbf x}^2_2\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \sum_{i \mathop = 1}^m \paren {\sum_{j \mathop = 1}^n a_{ij} x_j}^2\) | Definition of Euclidean Norm | |||||||||||
\(\ds \) | \(\le\) | \(\ds \sum_{i \mathop = 1}^m \paren {\sum_{j \mathop = 1}^n a_{ij}^2} \paren{\sum_{j \mathop = 1}^n x_j^2}\) | Cauchy's Inequality | |||||||||||
\(\ds \) | \(=\) | \(\ds \norm {\mathbf x}_2^2 \paren {\sum_{i \mathop = 1}^m \sum_{j \mathop = 1}^n a_{ij}^2}\) | Definition of Euclidean Norm |
Hence:
- $\ds \norm{T_A \mathbf x}_2 \le \norm{\mathbf x}_2 \sqrt {\sum_{i \mathop = 1}^m \sum_{j \mathop = 1}^n a_{ij}^2 }$
By Continuity of Linear Transformation between Normed Vector Spaces $T_A$ is continuous.
Hence, all linear transformations between finite-dimensional normed vector spaces are continuous.
$\blacksquare$
Sources
- 2017: Amol Sasane: A Friendly Approach to Functional Analysis ... (previous) ... (next): Chapter $\S 2.3$: The normed space $\map {CL} {X,Y}$