Equivalence of Definitions of Norm of Linear Functional

Theorem
Let $H$ be a Hilbert space, and let $L$ be a bounded linear functional on $H$.

Define the following norms of $L$:


 * $(1): \quad \left\|{L}\right\|_1 = \sup \left\{{\left|{Lh}\right|: \left\|{h}\right\| \le 1}\right\}$
 * $(2): \quad \left\|{L}\right\|_2 = \sup \left\{{\left|{Lh}\right|: \left\|{h}\right\| = 1}\right\}$
 * $(3): \quad \left\|{L}\right\|_3 = \displaystyle \sup \left\{{\dfrac {\left|{Lh}\right|} {\left\|{h}\right\|}: h \in H\setminus \left\{\mathbf 0 \right\}}\right\}$
 * $(4): \quad \left\|{L}\right\|_4 = \inf \left\{{c > 0: \forall h \in H: \left|{Lh}\right| \le c \left\|{h}\right\|}\right\}$

Then:
 * $\left\|{L}\right\|_1 = \left\|{L}\right\|_2 = \left\|{L}\right\|_3 = \left\|{L}\right\|_4$

Corollary
For all $h \in H$, the following inequality holds:
 * $\left|{Lh}\right| \le \left\|{L}\right\| \left\|{h}\right\|$

Proof of Theorem
We have:
 * $\displaystyle\left\{{ h \in H : \left\Vert{h}\right\Vert = 1 }\right\} \subseteq \left\{{ h \in H : \left\Vert{h}\right\Vert \le 1 }\right\} \subseteq H$

So it follows from the definition of the supremum that
 * $\left\Vert{L}\right\Vert_2 \le \left\Vert{L}\right\Vert_1 \le \left\Vert{L}\right\Vert_3$

Next we show that $\left\Vert{L}\right\Vert_2 = \left\Vert{L}\right\Vert_3$.

Therefore
 * $\left\Vert{L}\right\Vert_1 = \left\Vert{L}\right\Vert_2 = \left\Vert{L}\right\Vert_3$.

Moreover, if $\left|{Lh}\right| \le c \left\|{h}\right\|$ for all $h \in H\setminus\left\{\mathbf 0 \right\}$, then we have:
 * $\displaystyle\left\|{L}\right\|_3 = \sup \left\{{\dfrac {\left|{Lh}\right|} {\left\|{h}\right\|}: h \in H \setminus \left\{\mathbf 0 \right\}}\right\} \le c$

Taking the infimum over all such $c$ this reads:
 * $\left\Vert{L}\right\Vert_3 \le \left\Vert{L}\right\Vert_4$

Suppose that $c_0 := \left\Vert{L}\right\Vert_4 > \left\Vert{L}\right\Vert_3$.

Then by the definitions of these two norms, this means that there exists $\epsilon > 0$ such that for every $h \in H\setminus\left\{\mathbf 0 \right\}$:
 * $\displaystyle\frac{|Lh|}{\Vert h\Vert} + \epsilon \le c_0$

But this in turn implies that for every $h \in H\setminus\left\{\mathbf 0 \right\}$:
 * $|Lh| \leq c_0 \Vert h\Vert - \epsilon \Vert h\Vert = \left( c_0 - \epsilon \right) \Vert h\Vert$

This contradicts the fact that $c_0$ is the least such number satisfying this inequality.

Therefore:
 * $\left\Vert{L}\right\Vert_3 = \left\Vert{L}\right\Vert_4$

and the proof is complete.

Proof of Corollary
If $h = \mathbf 0$ there is nothing to prove.

Let $h \ne \mathbf 0$.

By the definition of the supremum:
 * $\dfrac{\left|{Lh}\right|} {\left\Vert{h}\right\Vert} \le \left\Vert{L}\right\Vert_3 = \left\Vert{L}\right\Vert$

whence:
 * $\left|{Lh}\right| \le \left\|{L}\right\| \left\|{h}\right\|$