For a normed space (X), an isometric isomorphism can be defined from (X) to its second dual space (X''), i.e. (J: X ightarrow X''), such that for all (x in X), (J(x) = J_x) with (J_x) being defined as (J_x(x') = x'(x) ; (forall x' in X')). This map (J) is called the evaluation map. When the range of (J) is equal to (X''), we say (X) is reflexive. In this post, we'll prove that
- the evaluation map (J) really maps an element in (X) to an element in (X'');
- (J) is an isometric isomorphism from (X) to (J(X)).
Part 1
To prove (J(x) = J_x in X'' (forall x in X)), we should show that (J_x) is both linear and continuous.
For the linearity of (J_x), let (x', y' in X') and (a, b in mathbb{K}). Due to the fact that (X') is itself a linear space with respect to operator addition and scalar product in the sense of point-wise evaluation at (x), we have
[
egin{aligned}
J_x(ax' + by') &= (ax' + by')(x) = a x'(x) + b y'(x) \
&= a J_x(x') + b J_x(y')
end{aligned}.
]
This proves (J_x) is linear and this linearity actually inherits from the linear structure of (X').
For the continuity of (J_x), we need to show it is a bounded functional.
Because (x' in X') is bounded, for all (x' in X'),
[
abs{J_x(x')} = abs{x'(x)} leq
orm{x'}_{X'} cdot
orm{x}_X.
]
We can see the norm of (J_x), i.e. ( orm{J_x}_{X''}) is bounded by ( orm{x}_X). Therefore, (J_x) is continuous. To sum up, we have (J_x in X'').
Part 2
Next, we shall prove (J) is isometric, viz. norm-preserving.
In the above, we've already shown that ( orm{J_x}_{X''} leq orm{x}_X). If we can further prove ( orm{J_x}_{X''} geq orm{x}_X) so that ( orm{J_x}_{X''} = orm{x}_X), (J) must be norm-preserving. The proof of this depends on whether we can find an (x') in (X'), such that
[
frac{abs{J_x(x')}}{
orm{x'}_{X'}} =
orm{x}_X,
]
which naturally leads to
[
orm{x}_X leq
orm{J_x}_{X''}.
]
Let (x_0) be arbitrarily selected from (X). We can define a functional (x') which at the moment can only be evaluated at (x_0) as (x'(x_0) = orm{x_0}_X). Then we extend the domain of (x') to the subspace (M) of (X) spanned by (x_0)
[
M = span{x_0} = {x = c x_0 vert c in mathbb{K}}
]
and for all (x = c x_0 in M), define
[
x'(x) = x'(c x_0) = c
orm{x_0}_X.
]
It is obvious that the extended (x') on (M) is linear. In addition, we have
[
abs{x'(x)} = abs{x'(c x_0)} = abs{c x'(x_0))} =
orm{c x_0}_X =
orm{x}_X,
]
which indicates that (x') is bounded and ( orm{x'}_{X'} = 1). Hence, (x') belongs to the dual space (M') of (M).
Next, by applying the Hahn-Banach theorem, we can extend the domain of (x') from the subspace (M) of (X) to the whole space (X), while preserving the norm ( orm{x'}_{X'} = 1). Therefore, for this specific (x' in X'),
[
frac{abs{J_{x_0}(x')}}{
orm{x'}_{X'}} = frac{abs{x'(x_0)}}{1} =
orm{x_0}_X,
]
so that
[
orm{x_0}_X leq
orm{J_{x_0}}_{X''} leq
orm{x_0}_X.
]
Because (x_0) is arbitrarily selected from (X), we've proved that (J: X ightarrow X'') is really an isometric map.
To prove (J) is an isomorphism between (X) and (J(X) subset X''), we should prove (J) preserves the linear structure from (X) to (X'') and is also an injective map. For the preservation of linear structure, it has already been verified during the proof of the linearity of (J_x) as above. To show (J) is injective, let (x_1, x_2 in X) and (x_1 eq x_2). For sure we can find an (x') in (X') such that (x'(x_1) eq x'(x_2)). Then for this (x'), we have (J_{x_1}(x') = x'(x_1)) is different from (J_{x_2}(x') = x'(x_2)), which indicates (J_{x_1} eq J_{x_2}). Hence (J) is injective.
Conclusions
Summarizing the above proof, we arrive at the conclusion that (J) is an isometric isomorphism between (X) and (J(X) subset X'').
Remark The key step in the above is during the proof of isometry, where a specific functional (x') is firstly defined at a single point (x_0 in X) with its value equal to ( orm{x_0}_X). Then its domain is extended to the span of (x_0) and further to the whole space (X) by using the Hahn-Banach theorem, which ensures the extension is both continuous and norm-preserving.