description/proof of that for infinite-dimensional normed vectors space, "dual" of basis is not necessarily basis for normed covectors (dual) space
Topics
About: vectors space
The table of contents of this article
Starting Context
- The reader knows a definition of normed covectors (dual) space of normed vectors space.
- The reader knows a definition of basis of module.
- The reader admits the proposition that for any module with any basis, the components set of any element with respect to the basis is unique.
- The reader admits the proposition that for any real or complex vectors space, taking any basis and the sum of the absolute components for each vector is a norm.
Target Context
- The reader will have a description and a proof of the proposition that for an infinite-dimensional normed vectors space, the "dual" of any basis is not necessarily a basis for the normed covectors (dual) space.
Orientation
There is a list of definitions discussed so far in this site.
There is a list of propositions discussed so far in this site.
Main Body
1: Structured Description
Here is the rules of Structured Description.
Entities:
\(F\): \(\in \{\mathbb{R}, \mathbb{C}\}\), with the canonical field structure
\(V\): \(\in \{\text{ the } F \text{ vectors spaces }\}\), with any norm
\(V^*\): \(= \text{ the normed covectors space of } V\)
\(B\): \(\in \{\text{ the bases for } V\}\), \(= \{b_j \vert j \in J\}\)
\(B^*\): \(= \{b^j \vert j \in J\}\) such that \(\forall j \in J (\forall l \in J (b^j (b_l) = \delta^j_l))\)
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Statements:
Not necessarily \(B^* \in \{\text{ the bases for } V^*\}\)
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2: Note
This is not claiming that \(B^*\) is never any basis.
Compare with the proposition that for any infinite-dimensional vectors space, the "dual" of any basis is never any basis for the covectors (dual) space: the proof of that has shown a covector that is not spanned by \(B^*\), but the issue is whether the covector is bounded.
This article uses the expression, "dual", with double-quotations because \(B^*\) is not usually called "dual" when it is not any dual basis.
3: Proof
Whole Strategy: Step 1: suppose that \(V\) has the norm that takes sum of absolute components; Step 2: see that \(B^*\) is well-defined; Step 3: take \(w \in V^*\) as for each \(v = v^{j_1} b_{j_1} + ... + v^{j_n} b_{j_n}\), \(w (v) = v^{j_1} + ... + v^{j_n}\), and see that \(w\) is bounded but not spanned by \(B^*\).
Step 1:
Let us suppose that \(V\) has the norm that takes sum of absolute components with respect to \(B\), which is possible, by the proposition that for any real or complex vectors space, taking any basis and the sum of the absolute components for each vector is a norm: this proposition is about that at least with respect to that norm, \(B^*\) is not any basis.
Step 2:
Let us see that \(B^*\) is well-defined.
Let \(v = v^{j_1} b_{j_1} + ... + v^{j_n} b_{j_n} \in V\) be any, which is the unique decomposition with nonzero components, by the proposition that for any module with any basis, the components set of any element with respect to the basis is unique.
\(b^j (v) = b^j (v^{j_1} b_{j_1} + ... + v^{j_n} b_{j_n}) = v^{j_1} b^j (b_{j_1}) + ... + v^{j_n} b^j (b_{j_n}) = v^{j_1} \delta^j_{j_1} + ... + v^{j_n} \delta^j_{j_n}\), determined uniquely.
\(b^j\) is linear, because for each \(v = v^{j_1} b_{j_1} + ... + v^{j_n} b_{j_n}, v' = v'^{l_1} b_{l_1} + ... + v'^{l_m} b_{l_m} \in V\) and each \(r, r' \in F\), when \(b_j\) is not contained in \(v\) nor \(v'\), \(b^j (r v + r' v') = 0 = 0 + 0 = r b^j (v) + r' b^j (v')\); when \(b_j\) is contained only in \(v\) as \(b_{j_p}\), \(b^j (r v + r' v') = r v^{j_p} = r v^{j_p} + 0 = r b^j (v) + r' b^j (v')\); when \(b_j\) is contained only in \(v'\) as \(b_{l_q}\), \(b^j (r v + r' v') = r' v'^{l_q} = 0 + r' v'^{l_q} = r b^j (v) + r' b^j (v')\); when \(b_j\) is contained both in \(v\) as \(b_{j_p}\) and in \(v'\) as \(b_{l_q}\), \(b^j (r v + r' v') = r v^{j_p} + r' v^{l_q} = r b^j (v) + r' b^j (v')\).
Let us see that \(b^j\) is bounded.
Let us think of \(sup_{v \in V \setminus \{0\}} \vert b^j (v) \vert / \Vert v \Vert\).
As \(v = v^1 b_{j_1} + ... v^m b_{j_m}\), \(\vert b^j (v) \vert / \Vert v \Vert = \vert b^j (v^1 b_{j_1} + ... v^m b_{j_m}) \vert / \Vert v^1 b_{j_1} + ... v^m b_{j_m} \Vert\).
When \(j \notin \{j_1, ..., j_m\}\), it is \(0\).
When \(j \in \{j_1, ..., j_m\}\) as \(j = j_l\), it is \(\vert v^l \vert / \Vert v^1 b_{j_1} + ... + v^m b_{j_m} \Vert\), but \(\Vert v^1 b_{j_1} + ... + v^m b_{j_m} \Vert = \vert v^1 \vert + ... + \vert v^m \vert\) by our choice of norm, so, \(\vert b^j (v) \vert / \Vert v \Vert = \vert v^l \vert / (\vert v^1 \vert + ... + \vert v^m \vert) \le 1\).
So, \(sup_{v \in V \setminus \{0\}} \vert b^j (v) \vert / \Vert v \Vert \le 1\), which means that \(b^j\) is bounded.
So, \(b^j \in V^*\).
Step 3:
Let us take \(w \in V^*\) as for each \(v = v^{j_1} b_{j_1} + ... + v^{j_n} b_{j_n}\), \(w (v) = v^{j_1} + ... + v^{j_n}\).
\(w\) is well-defined, because the decomposition is unique with nonzero components, as before.
Let us see that \(w\) is indeed linear (this is exactly the same as in Proof in the proposition that for any infinite-dimensional vectors space, the "dual" of any basis is never any basis for the covectors (dual) space).
Let \(v = v^{k_1} b_{k_1} + ... + v^{k_m} b_{k_m}, v' = v'^{l_1} b_{l_1} + ... + v'^{l_n} b_{l_n} \in V\) and \(r, r' \in F\) be any.
Let \(K := \{k_1, ..., k_m\}\) and \(L := \{l_1, ..., l_n\}\).
\(K \cap L\) may be empty or nonempty, but anyway, \(v = \sum_{k \in K \setminus L} v^k b_k + \sum_{k \in K \cap L} v^k b_k\) and \(v' = \sum_{l \in L \setminus K} v^l b_l + \sum_{k \in K \cap L} v^k b_k\).
\(r v + r' v' = r (\sum_{k \in K \setminus L} v^k b_k + \sum_{k \in K \cap L} v^k b_k) + r' (\sum_{l \in L \setminus K} v'^l b_l + \sum_{k \in K \cap L} v'^k b_k) = r \sum_{k \in K \setminus L} v^k b_k + r' \sum_{l \in L \setminus K} v'^l b_l + \sum_{k \in K \cap L} (r v^k + r' v'^k) b_k\).
So, \(w (r v + r' v') = r \sum_{k \in K \setminus L} v^k + r' \sum_{l \in L \setminus K} v'^l + \sum_{k \in K \cap L} (r v^k + r' v'^k) = r \sum_{k \in K \setminus L} v^k + \sum_{k \in K \cap L} r v^k + r' \sum_{l \in L \setminus K} v'^l + \sum_{k \in K \cap L} r' v'^k = r (\sum_{k \in K \setminus L} v^k + \sum_{k \in K \cap L} v^k) + r' (\sum_{l \in L \setminus K} v'^l + \sum_{k \in K \cap L} v'^k) = r \sum_{k \in K} v^k + r' \sum_{l \in L} v'^l = r w (v) + r' w (v')\).
Let us see that \(w\) is bounded.
Let us think of \(sup_{v \in V \setminus \{0\}} \vert w (v) \vert / \Vert v \Vert\).
As \(v = v^1 b_{j_1} + ... + v^m b_{j_m}\), \(\vert w (v) \vert / \Vert v \Vert = \vert w (v^1 b_{j_1} + ... + v^m b_{j_m}) \vert / \Vert v^1 b_{j_1} + ... + v^m b_{j_m} \Vert = \vert v^1 + ... + v^m \vert / (\vert v^1 \vert + ... + \vert v^m \vert) \le (\vert v^1 \vert + ... + \vert v^m \vert) / (\vert v^1 \vert + ... + \vert v^m \vert) = 1\).
So, \(sup_{v \in V \setminus \{0\}} \vert w (v) \vert / \Vert v \Vert \le 1\), which means that \(w\) is bounded.
So, \(w \in V^*\).
But \(w\) is not spanned by \(B^*\), because supposing that \(w = w^1 b^{j_1} + ... + w^m b^{j_m}\), there is a \(b^l \in B^* \setminus \{b^{j_1}, ..., b^{j_m}\}\), then, \(w (b_l) = 1\) but \((w^1 b^{j_1} + ... + w^m b^{j_m}) (b_l) = 0 + ... + 0 = 0\), a contradiction.
So, \(w\) is not spanned by \(B^*\).
So, \(B^*\) is not any basis.
