18.3 Conjugate Functions

Proof. Fix $\phi \in F$. Let $\alpha \in \real$ such that $\dpn{x, \phi}{\lambda}- \alpha \le f(x)$ for all $x \in E$, then for any $x \in E$,

so

On the other hand, suppose that $\alpha = \sup_{x \in E}\dpn{x, \phi}{\lambda}- f(x) < \infty$, then

for all $x \in E$. Therefore

$\square$

Proof. (1) $\Rightarrow$ (2): Let $\phi \in \bracsn{f^* < \infty}$, then by Definition 18.3.2, $(\phi, f^{*}(\phi)) \le f$.

(2) $\Rightarrow$ (1): By Definition 18.3.2, $f^{*}(\phi) \le \alpha$.$\square$

Proof. (1): By Proposition 18.1.6 and Proposition 5.22.3.

(3): For each $x \in E$ and $\phi \in F$,

As the above holds for all $\phi \in F$, $f^{**}\le f$.$\square$

Proof. Let $x \in E$ and $\phi \in F$. Assume without loss of generality that $f(x) < \infty$, then

Now let $x \in E$ and $\phi \in F$ such that $\dpn{x, \phi}{\lambda}= f(x) + f^{*}(\phi)$, then $f(x), f^{*}(\phi) < \infty$ and

By definition,

so for every $h \in E$,

Thus $\phi$ satisfies the subgradient inequality, and $\phi \in \partial f(x)$.

On the other hand, if $f(x) < \infty$ and $\phi \in \partial f(x)$, then $f(x + h) - f(x) \ge \dpn{h, \phi}{\lambda}$ for all $h \in E$, and

$\square$

Proof. Let

For each $(x, \alpha), (y, \beta) \in A$, $t \in [0, 1]$, and $\gamma \ge (1 - t)\alpha + t\beta$, there exists $\alpha' \ge \alpha$ and $\beta' \ge \beta$ such that $\gamma = (1 - t)\alpha' + t\beta'$. In which case,

so $\bracs{(1 - t)x + ty}\times [\gamma, \infty) \subset \ol{\text{Conv}}(\text{epi}(f))$, $(1 - t)(x + \alpha) + t(y, \beta) \in A$, and $A$ is convex.

Let $(x, \alpha) \in \ol A$, then there exists a net $\langle (x_{\gamma}, \alpha_{\gamma}) \rangle_{\gamma \in C}\subset A$ with $(x_{\gamma}, \alpha_{\gamma}) \to (x, \alpha)$. In which case, for each $r > 0$, $\langle (x_{\gamma}, \alpha_{\gamma} + r) \rangle_{\gamma \in C}\subset A$ and $(x_{\gamma}, \alpha_{\gamma} + r) \to (x, \alpha + r)$, so $(x, \alpha + r) \in \ol{\text{Conv}}(\text{epi}(f))$ and

Thus $(x, \alpha) \in A$ and $A$ is closed.

Since $A$ is a closed convex set containing $\text{epi}(f)$, $A = \ol{\text{Conv}}(\text{epi}(f))$.$\square$

Proof. (1): Since $f$ is convex and $\sigma(E, F)$-lower semicontinuous, $\text{epi}(f)$ is $\sigma(E \times \real, F \times \real)$-closed and convex. By the Hahn-Banach Theorem, there exists $\phi \in F$ and $\mu \in \real$ such that

For any $(y, \beta) \in \text{epi}(f)$, $\beta$ may be arbitrarily large, $\mu \ge 0$. In particular, since $(x, f(x)) \in \text{epi}(f)$, the strict inequality implies that $\mu > 0$. Thus for each $y \in \bracs{f < \infty}$,

so $(\mu^{-1}\phi, \dpn{x, \mu^{-1}\phi}{E}- \alpha) \le f$ and

$\square$

Proof. (1): Let $\phi \in F$ such that $f^{*}(\phi) < \infty$, then for each $x \in E$,

so $\dpn{\cdot, \phi}{\lambda}- f^{*}(\phi) \le f$, and

On the other hand, let $\phi \in F$ and $\alpha \in \real$ such that $(\phi, \alpha) \le f$, then $f^{*}(\phi) \le \alpha$, and

Therefore

(2): By Lemma 18.3.4, $f^{**}$ is lower semicontinuous and convex with $f^{**}\le f$, so $\text{epi}(f^{**}) \supset \text{epi}(f)$ and $\text{epi}(f^{**}) \supset \ol{\text{Conv}}(\text{epi}(f))$. Thus it is sufficient to show that $\text{epi}(f^{**}) \subset \ol{\text{Conv}}(\text{epi}(f))$, or equivalently,

To this end, let $A = \ol{\text{Conv}}(\text{epi}(f))$ and $(x, \alpha) \in E \times \real \setminus A$. By the Hahn-Banach Theorem, there exists $\phi \in F$, $\mu \in \real$, and $\alpha_{0} \in (\alpha, \infty)$ such that

For any $(y, \beta) \in A$, $\beta$ may be arbitrarily large by Lemma 18.3.6, so $\mu \ge 0$.

In the case that $\mu > 0$, for each $y \in \bracs{f < \infty}$,

so $(\mu^{-1}\phi, \dpn{x, \mu^{-1}\phi}{\lambda}- \alpha_{0}) \le f$ and

Now suppose that $\mu = 0$. Given that $f^{*} \ne \infty$, there exists at least one pair $(\phi_{0}, \gamma_{0}) \in F \times \real$ such that $(\phi_{0}, \gamma_{0}) \le f$. Let

For each $t > 0$, let $\Phi_{t} = \phi_{0} + t\phi$ and $\Gamma_{t} = \gamma_{0} + t\gamma$, then for each $y \in \bracs{f < \infty}$,

so $(\Phi_{t}, \Gamma_{t}) \le f$. By (1),

As the above holds for all $t > 0$, $f^{**}(x) = \infty > \alpha$.

Thus $f^{**}(x) > \alpha$ and $(x, \alpha) \not\in \text{epi}(f^{**})$ for all $(x, \alpha) \in E \times \real \setminus A$. Therefore

and $\text{epi}(f^{**}) \subset \ol{\text{Conv}}(\text{epi}(f))$.$\square$

Proof. For each $(\phi, \alpha) \in F \times \real$, denote $(\phi, \alpha) \le f$ if $\dpn{\cdot, \phi}{\lambda}- \alpha \le f$. By the Fenchel-Moreau Theorem,

for all $x \in E$. By Proposition 13.4.1,

is separable with respect to $\sigma(F \times \real, E \times \real)$. Therefore there exists $\seq{(\phi_n, \alpha_n)}\subset S$ such that for each $x \in E$,

$\square$