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    Linear functionals: definition, boundedness equivalent to continuity, and completeness of the dual

    P-MNEPM

    Area
    Real Analysis
    Topics
    Continuity, Dual Spaces, Functional Analysis, Norms
    Status
    draft

    Let \(X\) be a normed vector space.

    1. Give the definition of what it means for a map \(L:X\to \CC\) to be a linear functional.

    2. Define what it means for \(L\) to be bounded and show \(L\) is bounded \(\iff L\) is continuous.

    3. Prove that \((\dualof{X}, \norm{\wait}_{\operatorname{op}})\) is a Banach space.

    Solution

    1. \(L\colon X \to \CC\) is a linear functional if \(L(x + y) = L(x) + L(y)\) and \(L(\alpha x) = \alpha L(x)\) for all \(x, y \in X\) and \(\alpha \in \CC\).

    2. A linear functional \(L\) is bounded if \(\|L\|_{\operatorname{op}} \coloneqq \sup_{\|x\| \le 1}|L(x)| < \infty\), equivalently if \(|L(x)| \le C\|x\|\) for some \(C\) and all \(x\).

    1. A linear functional \(L\) is bounded if and only if it is continuous.

    1.1. If \(L\) is bounded, then \(L\) is continuous.

    \(|L(x) - L(y)| = |L(x - y)| \le \|L\|_{\operatorname{op}}\|x - y\|\).

    1.2. If \(L\) is continuous, then \(L\) is bounded.

    Suppose not. Then there are \(x_n\) with \(\|x_n\| = 1\) and \(|L(x_n)| \ge n\). Then \(y_n = x_n/n \to 0\) while \(|L(y_n)| \ge 1\), contradicting continuity at \(0\), where \(L(0) = 0\).

    1.3. Q.E.D.

    Steps 1 and 2.

    1. Let \(X^*\) be the space of bounded linear functionals on \(X\).

    2. \(\|\cdot\|_{\operatorname{op}}\) is a norm on \(X^*\).

    If \(\|L\|_{\operatorname{op}} = 0\) then \(L(x) = 0\) for \(\|x\| \le 1\), hence for all \(x\) by scaling. \(\|\alpha L\|_{\operatorname{op}} = |\alpha|\,\|L\|_{\operatorname{op}}\), and \(|(L + M)(x)| \le |L(x)| + |M(x)|\) gives \(\|L + M\|_{\operatorname{op}} \le \|L\|_{\operatorname{op}} + \|M\|_{\operatorname{op}}\).

    3. Every Cauchy sequence \((L_n)\) in \(X^*\) converges in \(X^*\).

    3.1. For each \(x \in X\), \(L(x) \coloneqq \lim_n L_n(x)\) exists, and \(L\) is linear.

    \(|L_n(x) - L_m(x)| \le \|L_n - L_m\|_{\operatorname{op}}\|x\|\), so \((L_n(x))\) is Cauchy in the complete space \(\CC\). Linearity passes to the limit: \(L(\alpha x + y) = \lim_n(\alpha L_n(x) + L_n(y)) = \alpha L(x) + L(y)\).

    3.2. Q.E.D.

    Given \(\eps > 0\), choose \(N\) with \(\|L_n - L_m\|_{\operatorname{op}} < \eps\) for \(n, m \ge N\). For \(\|x\| \le 1\) and \(n \geq N\), \(|L(x) - L_n(x)| = \lim_m |L_m(x) - L_n(x)| \le \eps\), so \(\|L - L_n\|_{\operatorname{op}} \le \eps\). Then \(\|L\|_{\operatorname{op}} \le \|L_N\|_{\operatorname{op}} + \eps\), so \(L \in X^*\), and \(L_n \to L\) in \(X^*\).

    4. Q.E.D.

    Steps 2 and 3 show that \(X^*\) is a Banach space.

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