# 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

- 1990: John B. Conway:
*A Course in Functional Analysis*(2nd ed.) ... (previous) ... (next): $\text{I}$ Hilbert Spaces: $\S 3.$ The Riesz Representation Theorem: Proposition $3.3$