Supremum Operator Norm of Linear Transformation is Bounded Above by Hilbert-Schmidt Norm
Theorem
Let $T_A : \R^n \to \R^m$ be the linear transformation such that:
- $\forall \mathbf x \in \R^n : T_A \mathbf x := A \mathbf x$
Let $\norm {\, \cdot \,}$ be the supremum operator norm.
Let $\norm {\, \cdot \,}_{HS}$ be the Hilbert Schmidt norm.
Then:
- $\norm {T_A} \le \norm A_{HS}$
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:
- $\forall \mathbf x \in \R^n : T_A \mathbf x := A \mathbf x$
From Linear Transformations between Finite-Dimensional Normed Vector Spaces are Continuous:
- $\ds \norm{T_A \mathbf x}_2 \le \norm{\mathbf x}_2 \sqrt {\sum_{i \mathop = 1}^m \sum_{j \mathop = 1}^n a_{ij}^2 }$
Take the supremum of both sides with the condition $\forall \mathbf x \in \R^n : \norm {\mathbf x}_2 \le 1$.
On the left we are left with $\norm {T_A}$.
On the right we have:
\(\ds \norm {\norm{\mathbf x}_2 \sqrt {\sum_{i \mathop = 1}^m \sum_{j \mathop = 1}^n a_{ij}^2 } }\) | \(=\) | \(\ds \sup \set {\norm{\mathbf x}_2 \sqrt {\sum_{i \mathop = 1}^m \sum_{j \mathop = 1}^n a_{ij}^2 } : \mathbf x \in \R^n : \norm {\mathbf x}_2 \le 1 }\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \sqrt {\sum_{i \mathop = 1}^m \sum_{j \mathop = 1}^n a_{ij}^2 } \sup \set {\norm{\mathbf x}_2 : \mathbf x \in \R^n : \norm {\mathbf x}_2 \le 1 }\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \sqrt {\sum_{i \mathop = 1}^m \sum_{j \mathop = 1}^n a_{ij}^2 } \cdot 1\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \norm {A}_{HS}\) | Definition of Hilbert-Schmidt Norm |
Hence:
- $\norm {T_A} \le \norm {A}_{HS}$
$\blacksquare$
Sources
- 2017: Amol Sasane: A Friendly Approach to Functional Analysis ... (previous) ... (next): Chapter $\S 2.3$: The normed space $\map {CL} {X,Y}$. Operator norm and the normed space $\map {CL} {X, Y}$