Theorem 18.4.4.label Let $\dpn{E, F}{\lambda}$ be a duality over $\real$, $f: X \to (-\infty, \infty]$ be a $\sigma(E, F)$-lower semicontinuous, subadditive, and positively homogeneous function with $f(0) = 0$, and

\[\Sigma = \bracs{\phi \in F| (\phi, 0) \le f}\]

then:

  1. (1)

    $f^{*} = I_{\Sigma}$.

  2. (2)

    $\Sigma$ is the unique non-empty $\sigma(F, E)$-closed convex subset of $F$ such that $f = H_{\Sigma}$.

  3. (3)

    $\Sigma$ is equicontinuous if and only if there exists $U \in \cn_{E}(0)$ such that $\sup_{x \in U}f(x) < \infty$.

Conversely,

  1. (3)

    For any non-empty $\Sigma \subset A$, $H_{\Sigma}$ is a lower semicontinuous, subadditive, and positively homogeneous function with $H_{\Sigma}(0) = 0$.

Proof, [Theorem 4.25, Cla13]. (1): Let $\phi \in \Sigma$, then since $\dpn{x, \phi}{\lambda}\le f(x)$ for all $x \in E$,

\[0 \ge f^{*}(\phi) = \sup_{x \in E}\dpn{x, \phi}{\lambda}- f(x) = \dpn{0, \phi}{\lambda}- f(0) = 0\]

where the supremum is achieved at $0$. On the other hand, if $\phi \in F \setminus \Sigma$, then there exists $x \in E$ such that $\dpn{x, \phi}{\lambda}- f(x) > 0$. In which case, by positive homogeneity of $f$,

\[f^{*}(\phi) \ge \sup_{\mu > 0}\dpn{\mu x, \phi}{\lambda}- f(\mu x) = \sup_{\mu > 0}\mu\braks{\dpn{ x, \phi} - f(x)}= \infty\]

Thus $f^{*}= I_{\Sigma}$.

(2): By Lemma 18.3.7, $I_{\Sigma} \ne \infty$, so $\Sigma \ne \emptyset$. Since

\[\Sigma = \bigcap_{x \in E}\bracs{\phi \in F| \dpn{x, \phi}{\lambda} \le f(x)}\]

is an intersection of $\sigma(F, E)$-closed and convex sets, it is also $\sigma(F, E)$-closed.

Now, let $x, y \in E$ and $t \in [0, 1]$, then since $f$ is subadditive and positively homogeneous,

\[f((1 - t)x + ty) \le f((1 - t)x) + f(ty) = (1 - t)f(x) + tf(y)\]

so $f$ is convex. Given that $f$ is also $\sigma(E, F)$-lower semicontinuous, the Fenchel-Moreau Theorem and (4) of Lemma 18.4.3 imply that

\[f = f^{**}= I_{\Sigma}^{*} = H_{\Sigma}\]

By (2) of Lemma 18.4.3, $\Sigma$ is the unique closed $\sigma(F, E)$-convex set such that $f = H_{\Sigma}$.

(3): Let $U \in \cn_{E}(0)$ be circled such that $M = \sup_{x \in U}f(x) < \infty$, then

\[\sup_{\substack{y \in \Sigma \\ x \in U}}\dpn{x, y}{\lambda}\le \sup_{x \in U}f(x) = M < \infty\]

Since $U$ is circled, $\bigcup_{y \in \Sigma}\dpn{M^{-1}U, y}{\lambda}\subset \ol{B_\real(0, 1)}$, so $\Sigma$ is equicontinuous by Proposition 11.14.1.

Conversely, if $\Sigma$ is equicontinuous, then there exists $U \in \cn_{E}(0)$ such that $M = \sup_{y \in \Sigma, x \in U}\dpn{x, y}{\lambda}< \infty$. In which case, (2) implies that

\[\sup_{x \in U}f(x) = \sup_{x \in U}H_{\Sigma}(x) = \sup_{\substack{y \in \Sigma \\ x \in U}}\dpn{x, y}{\lambda}= M < \infty\]

(4): By Proposition 5.22.3, $H_{\Sigma}$ is lower semicontinuous.

Let $x, y \in E$ and $\mu > 0$, then

\begin{align*}H_{\Sigma}(x + y)&= \sup_{z \in \Sigma}\dpn{x + y, z}{\lambda}\\&\le \sup_{z \in \Sigma}\dpn{x, z}{\lambda}+ \sup_{z \in \Sigma}\dpn{y, z}{\lambda}= H_{\Sigma}(x) + H_{\Sigma}(y)\end{align*}

and

\[H_{\Sigma}(\mu x) = \sup_{z \in \Sigma}\dpn{\mu x, z}{\lambda}= \mu\sup_{z \in \Sigma}\dpn{x, z}{\lambda}= \mu H_{\Sigma}(x)\]

so $H_{\Sigma}$ is subadditive and positively homogeneous.$\square$

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