# 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

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:

$A = \begin {bmatrix} a_{1 1} & \cdots & a_{1 n} \\ \vdots & \ddots & \vdots \\ a_{m 1} & \cdots & a_{m n} \\ \end{bmatrix}$

Let $T_A : \R^n \to \R^m$ be the linear transformation such that:

$\forall \mathbf x \in \R^n : T_A \mathbf x := A \mathbf x$
$\ds \norm{T_A \mathbf x}_2 \le \norm{\mathbf x}_2 \sqrt {\sum_{i \mathop = 1}^m \sum_{j \mathop = 1}^n a_{ij}^2 }$

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:

$\norm {T_A} \le \norm {A}_{HS}$

$\blacksquare$