Equivalence of Definitions of Norm of Linear Transformation/Definition 1 Greater or Equal Definition 3

Theorem
Let $H, K$ be Hilbert spaces.

Let $A: H \to K$ be a bounded linear transformation.

Let:
 * $\lambda_1 = \sup \set {\norm {A h}_K: \norm h_H \le 1}$

and
 * $\lambda_3 = \sup \set {\norm {A h}_K: \norm h_H = 1}$

Then:
 * $\lambda_1 \ge \lambda_3$

Proof
By definition of the supremum:
 * $\forall h \in H, \norm h_H \le 1 : \norm{A h}_K \le \lambda_1$

In particular:
 * $\forall h \in H, \norm h_H = 1 : \norm{A h}_K \le \lambda_1$

From Continuum Property:
 * $\lambda_3 = \sup \set {\norm {A h}_K: \norm h_H = 1}$ exists

By definition of the supremum:
 * $\lambda_3 \le \lambda_1$