Equivalence of Definitions of Norm of Linear Functional

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Theorem

Let $V$ be a normed vector space, and let $L$ be a bounded linear functional on $V$.

Define the following norms of $L$:

$(1): \quad \norm L_1 = \sup \set {\size {L h}: \norm h \le 1}$
$(2): \quad \norm L_2 = \sup \set {\size {L h}: \norm h = 1}$
$(3): \quad \norm L_3 = \sup \set {\dfrac {\size {L h} } {\norm h}: h \in H \setminus \set {\mathbf 0} }$
$(4): \quad \norm L_4 = \inf \set {c > 0: \forall h \in H: \size {L h} \le c \norm h}$

Then:

$\norm L_1 = \norm L_2 = \norm L_3 = \norm L_4$


Corollary

For all $v \in V$, the following inequality holds:

$\size {L v} \le \norm L \norm v$


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Proof

We have:

$\set {v \in V : \norm v = 1} \subseteq \set {v \in V : \norm v \le 1} \subseteq V$

So it follows from the definition of the supremum that

$\norm L_2 \le \norm L_1 \le \norm L_3$

Next we show that $\norm L_2 = \norm L_3$:

\(\ds \norm L_2\) \(=\) \(\ds \sup \set {\norm {L v}: \norm h = 1}\)
\(\ds \) \(=\) \(\ds \sup \set {\norm {L v'}: v \in V \setminus \set {\mathbf 0}, \ v' = v / \norm v}\) as $\norm {v / \norm v} = \norm v / \norm v = 1$ for all $v \in V \setminus \set {\mathbf 0}$
\(\ds \) \(=\) \(\ds \sup \set {\norm {\frac 1 {\norm v} L v}: v \in V \setminus \set {\mathbf 0} }\) as $L$ is linear
\(\ds \) \(=\) \(\ds \sup \set {\frac {\norm {L v} } {\norm v}: v \in V \setminus \set {\mathbf 0} }\) Norm Axiom $\text N 2$: Positive Homogeneity
\(\ds \) \(=\) \(\ds \norm L_3\)

Therefore

$\norm L_1 = \norm L_2 = \norm L_3$

Moreover, if $\size {L v} \le c \norm v$ for all $v \in V \setminus \set {\mathbf 0}$, then we have:

$\norm L_3 = \sup \set {\dfrac {\size {L v} } {\norm v}: v \in V \setminus \set {\mathbf 0} } \le c$

Taking the infimum over all such $c$ this reads:

$\norm L_3 \le \norm L_4$

Aiming for a contradiction, suppose $c_0 := \norm L_4 > \norm L_3$.

Then by the definitions of these two norms, this means that there exists $\epsilon > 0$ such that for every $v \in V \setminus \set {\mathbf 0}$:

$\dfrac {\size {L v} } {\norm v} + \epsilon \le c_0$

But this in turn implies that for every $v \in V \setminus \set {\mathbf 0}$:

$\size {L v} \le c_0 \norm v - \epsilon \norm v = \paren {c_0 - \epsilon} \norm v$

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

Therefore:

$\norm L_3 = \norm L_4$

and the proof is complete.

$\blacksquare$


Sources

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