# Solutions to infra-exercises I

4 September 2022

These are S’s solutions for the first set of infra-Bayesian exercises. Spoilers, of course!

I decree this a Moore method problem set! That means you’re allowed to be as pedantic about definitions as you like, and I commit to addressing your pedantry. Please shoot me an email if there’s a way I can improve this document’s rigor and/or clarity!

When the document says monotone, it means what most analysis textbooks call increasing. A better term in the context of posets, where the document defines monotonicity, would be order-preserving.

I shall use the terminology 0-nonnegative rather than 0-increasing for a functional $$f : V \to \mathbb R$$ with $$f(\mathbf 0) \geq 0,$$ since I find the latter misleading. I also like the terminology non-expansive for 1-Lipschitzness, since it describes the geometric intuition rather than being named after a dude, but won’t use it here for clarity.

I use the variable $$t$$ for mixture coefficients out of habit.It’s typically used in topology, for instance when defining homotopy.

## Problem 1

Let $$V$$ be a real vector space. Show that a function $$g : V \to \mathbb R$$ is affine iff $$g(x) = f(x) + c,$$ where $$f(x) : V \to W$$ is linear and $$c \in \mathbb R.$$

### Solution

If: Let $$g(x)$$ be as given. For any $$v_1, v_2 \in V$$ and $$t \in [0, 1],$$ we have \begin{align*} g(tv_1 + (1 - t)v_2) &= f(tv_1 + (1 - t)v_2) + c \newline &= tf(v_1) + (1 - t)f(v_2) + c \newline &= t(f(v_1) + c) + (1 - t)(f(v_2) + c) \newline &= t\cdot g(v_1) + (1 - t)\cdot g(v_2). \end{align*}

Only if: Let $$g(x)$$ be affine. Define $$f(x) = g(x) - g(\mathbf 0).$$ We must show that $$f(x)$$ is linear; i.e., homogeneous and additive.

To show homogeneity, we have for any $$c \in [0, 1]$$ \begin{align*} f(cx) &= f(cx + (1 - c)\mathbf 0) \newline &= g(cx + (1 - c)\mathbf 0) - g(\mathbf 0) \newline &= c\cdot g(x) + (1 - c)\cdot g(\mathbf 0) - g(\mathbf 0) \newline &= c\cdot g(x) - g(\mathbf 0) = c \cdot f(x). \end{align*}

For $$c > 1$$ we have $$c^{-1} \in (0, 1).$$ Therefore \begin{align*} f(cx) &= f(c^2x/c) = \dfrac1c f(c^2 x). \end{align*}

Rewriting, we have $$cf(cx) = f(c^2x).$$ Setting $$y = cx,$$ we obtain the desired. It remains to show that $$f$$ is homogeneous for a negative constant. We have \begin{align*} f(-x) &= g(-x) - g(\mathbf 0) \newline &= g(-x) - g\left(\dfrac12 x + \dfrac12 (-x)\right) \newline &= g(-x) - \dfrac12 g(x) - \dfrac12 g(-x) \newline &= \dfrac12 g(-x) - \dfrac12g(x) \newline &= \dfrac12 f(-x) - \dfrac12f(x), \end{align*} which implies $$f(-x) = -f(x).$$

To show additivity, we have for any $$v_1, v_2 \in V$$ \begin{align*} f(v_1 + v_2) &= 2f\left(\dfrac12 v_1 + \dfrac12 v_2\right) \newline &= 2g\left(\dfrac12 v_1 + \dfrac12 v_2 \right) - 2g(\mathbf 0) \newline &= g(v_1) + g(v_2) - 2g(\mathbf 0) = f(v_1) + f(v_2). \end{align*}

This completes the proof.

## Problem 2

Let $$f : \mathbb R \to \mathbb R$$ be continuous and concave. Let $$T$$ be the set of all affine functions $$\phi : \mathbb R \to \mathbb R$$ such that $$\phi(x) \geq f(x)$$ for all $$x.$$ Show that $$f(x) = \inf T(x).$$

### Solution

Fix an $$a \in \mathbb R.$$ Assume $$f(a)$$ is defined. $$f(a)$$ is a lower bound for $$T(a)$$ by construction.

Suppose $$k > f(a).$$ We will show that there is a $$\phi \in T$$ with $$k > \phi(a).$$

We claim the set $$C = \{(x, y) \in \mathbb R^2 \mid y \leq f(x)\}$$ is convex and closed:

• Convex: Follows directly from concavity of $$f.$$
• Closed: We show that the complement of $$C$$ $C' = \{(x, y) \in \mathbb R^2 \mid y > f(x)\}$ is open under the $$\infty$$-norm. Let $$p = (x, y) \in C'.$$ Let $$\varepsilon_1 = (y - f(x))/2 > 0.$$ By continuity of $$f$$ there is a $$\delta > 0$$ such that $|x_1 - x_2| < \delta \implies |f(x_1) - f(x_2)| < \varepsilon_1.$ Let $$\varepsilon = \min \{\varepsilon_1, \delta\}.$$ Then whenever $$|x - x_1| < \varepsilon$$ and $$|y - y_1| < \varepsilon,$$ we have \begin{align*} y_1 - f(x_1) &= (y_1 - y) + \left(y - f(x)\right) + \left(f(x) - f(x_1)\right) \newline &> -\varepsilon + 2\varepsilon_1 - \varepsilon_1 = \varepsilon_1 - \varepsilon \geq 0. \end{align*}

Since $$v = (a, k) \in \mathbb R^2 \setminus C,$$ the Hahn-Banach separation theorem furnishes a linear, continuous $$F : \mathbb R^2 \to \mathbb R$$ and $$s \in \mathbb R$$ such that $$F(c) \leq s < F(v)$$ for any $$c \in C.$$ In particular, there are constants $$m, n \in \mathbb R$$ such that $$F((x, y)) = mx + ny.$$ The Hahn-Banach condition then says $mx + n \cdot f(x) \leq s < ma + nk.$

If $$n = 0,$$ we have $$mx < ma$$ for all $$(x, y) \in C,$$ which is absurd, since $$(a, f(a)) \in C.$$ Thus we may define $$\phi(x) = (s - mx)/n$$. Then we have that $$\phi$$ is affine, $$\phi(x) \geq f(x),$$ and $$\phi(a) < k.$$

This completes the proof.

## Problem 3a

Let $$f: \mathbb R \to \mathbb R$$ be concave, monotone, 0-nonnegative, and 1-Lipschitz. Let $$T$$ be the set of all affine functions $$\phi : \mathbb R \to \mathbb R$$ satisfying the latter three properties and $$\phi(x) \geq f(x).$$ Show that $$f(x) = \inf T(x).$$

### Solution

Construct $$\phi$$ as in the previous solution. We must show that $$\phi$$ is monotone, 0-nonnegative, and 1-Lipschitz.

• Monotonicity: Suppose contrariwise that $$x_1 < x_2$$ but $$\phi(x_1) > \phi(x_2).$$ Since $$\phi$$ is affine, there is some $$x_3 > x_1$$ with $$\phi(x_3) < f(x_1) < f(x_3),$$ a contradiction.
• 1-Lipschitz: Suppose contrariwise there are $$x_1, x_2 \in \mathbb R$$ and $$\varepsilon > 0$$ such that $\frac{\phi(x_2) - \phi(x_1)}{x_2 - x_1} = 1 + \varepsilon.$ Since $$\phi$$ is affine, $$\phi(x) = (1 + \varepsilon)x + \phi(0).$$ Let $c < \frac{f(0) - \phi(0)}{\varepsilon} < 0.$ Then we have $$f(c) \geq f(0) + c > \phi(0) + (1 + \varepsilon)c = \phi(c),$$ a contradiction.
• 0-nonnegativity: $$\phi(0) \geq f(0) \geq 0.$$

This completes the proof.

## Problem 3b

Let $$\Phi = \{\phi_i\}_{i \in I}$$ be a collection of affine functions mapping $$\mathbb R \to \mathbb R$$ that are each monotone, 1-Lipschitz, and 0-nonnegative. Show that $$f(x) = \inf \Phi(x)$$ is continuous, concave, monotone, 1-Lipschitz, and 0-nonnegative.

### Solution

• 0-nonnegativity: $$f(0) = \inf \Phi(0) = C.$$ Suppose $$C < 0.$$ Then there would exist $$\phi \in \Phi$$ with $$\phi(0) = C/2 < 0,$$ a contradiction.

• Monotonicity: Suppose contrariwise that $$x < y$$ but $$f(x) > f(y).$$ Then, for any $$\varepsilon > 0,$$ there is a $$\phi \in \Phi$$ with $$f(y) = \phi(y) + \varepsilon.$$ We have \begin{align*} f(x) - \phi(x) &> f(y) - \phi(x) \newline &\geq f(y) - [\phi(y) - (y - x)] \newline &= \varepsilon + (y - x) > 0, \end{align*} a contradiction.

• 1-Lipschitz: Suppose towards a contradiction that $$f$$ is not 1-Lipschitz. Then there exist $$x, y \in \mathbb R$$ with $\frac{|f(y) - f(x)|}{|y - x|} = L > 1.$ Suppose without loss of generality that $$x < y.$$ Let $$\varepsilon > 0,$$ choosing $$\varepsilon < (L - 1)|y - x|.$$ Since $$f(x) = \inf \Phi(x),$$ there is a $$\phi_i \in \Phi$$ such that $$\phi_i(x) = f(x) + \varepsilon.$$ Then we have \begin{align*} f(y) - \phi_i(y) &\geq f(y) - \left(\phi_i(x) + |y - x|\right) \newline &= f(y) - \left(f(x) + \varepsilon + |y - x|\right) \newline &= [f(y) - f(x)] - \varepsilon - |y - x| \newline &= L|y - x| - \varepsilon - |y - x| \newline &= (L - 1)|y - x| - \varepsilon > 0, \end{align*} a contradiction.

• Continuity: Follows immediately.

• Concavity: Suppose contrariwise that $$f$$ is not concave. Then there exist $$x, y \in \mathbb R$$ and $$t \in (0, 1)$$ such thatThe variable names here are for “line” and “curve." $L = t\cdot f(x) + (1 - t)f(y) > f(tx + (1 - t)y) = C.$

By assumption, there is a function $$\phi \in \Phi$$ such that $$\phi(tx + (1-t)y) = (L + C)/2.$$ We must have $$\phi(x) \geq f(x)$$ and $$\phi(y) \geq f(y).$$ But since $$\phi$$ is affine, we must have $$\phi(tx + (1-t)y) \geq L,$$ a contradiction.

## Problem 4a

Let $$\psi : C(X) \to \mathbb R$$ be concave, monotone, 1-Lipschitz, and 0-nonnegative. Let $$\Phi$$ be the set of all affine, monotone, 1-Lipschitz, and 0-nonnegative functions $$\phi : C(X) \to \mathbb R$$ such that $$\phi \geq \psi.$$ Show that $$\psi = \inf \Phi.$$

### Solution

$$\psi$$ is a lower bound for $$\Phi$$ by construction. Fix $$f \in C(X)$$ and suppose that $$k > \psi(f).$$ We shall produce a $$\phi \in \Phi$$ such that $$\psi(f) \leq \phi(f) < k.$$

We claim that the set $$S = \bigcup_{g \in C(X)} \left(g \times (-\infty, \psi(g)]\right) \subset C(X) \times \mathbb R$$ is closed and convex:

• Closed: We show that the complement of $$S$$ $S' = \bigcup_{g \in C(X)} \left(g \times (\psi(g), \infty)\right)$ is open under the $$\infty$$-norm. Let $$(g, c) \in S'.$$ Let $$\varepsilon = \left(c - \psi(g)\right)/2.$$ Then for any $$(h, k) \in C(X) \times \mathbb R$$ with $$d_\infty(g, h) < \varepsilon$$ and $$|c - k| < \varepsilon,$$ we have \begin{align*} k - \psi(h) &= (k - c) + \left(c - \psi(g)\right) + \left((\psi(g) - \psi(h)\right) \newline &> -\varepsilon + 2\varepsilon - d_\infty(g, h) \newline &> -\varepsilon + 2\varepsilon - \varepsilon = 0. \end{align*}
• Convex: Let $$(f_1, c_1), (f_2, c_2) \in S.$$ Let $$t \in (0, 1).$$ Then $$t \cdot f_1 + (1 - t)f_2 \in C(X),$$ and since $$\psi$$ is concave, $t \cdot \psi(f_1) + (1 - t)\psi(f_2) \leq \psi(t\cdot f_1 + (1-t) f_2).$

Therefore Hahn-Banach produces a linear functional $$F: C(X) \times \mathbb R \to \mathbb R$$ and $$s \in \mathbb R$$ such that for any $$(g, c) \in S,$$ $$F(g, c) \leq s < F(f, k).$$ Since $$F$$ is (bi)linear, we can write $F(g, c) = F(g, 0) + F(0, c) = M(g) + nc$ for some linear $$M : C(X) \to \mathbb R$$ and $$n \in \mathbb R.$$ The Hahn-Banach condition then says

$M(g) + nc \leq s < M(f) + nk.$

As before, $$M \equiv 0$$ and $$n = 0$$ are impossible. If $$M \not \equiv 0$$ and $$n = 0,$$ the Hahn-Banach condition is $$M(g) < M(f)$$ for all $$g \in C(X),$$ which is absurd. Thus, we may define $\phi(f) = \frac{s - M(g)}n.$

Note that $$\phi$$ is affine. Applying the Hahn-Banach condition to $$(f, \psi(f))$$ gives $$\psi(f) \leq \phi(f) < k,$$ as desired.

It remains to show that $$\phi$$ is monotone, 1-Lipschitz, and 0-nonnegative.

• 0-nonnegativity: $$\phi(0) \geq \psi(0) \geq 0.$$
• Monotonicity: Suppose contrariwise that $$f < g$$ but $$\phi(f) > \phi(g).$$ Since $$\phi$$ is affine, there is some $$c \in \mathbb R$$ and $$h(x) = f(x) + c$$ for which $$\phi(h) < \psi(f) < \psi(h),$$ a contradiction.
• 1-Lipschitz: This is the hard part.It took me multiple days by itself. Suppose there are $$f, g \in C(X)$$ and $$\varepsilon > 0$$ for which $|\phi(f) - \phi(g)| = (1 + \varepsilon) d_\infty(f,g).$ Let $$U = \max\{f,g\}, L = \min\{f,g\}.$$ Clearly $$U, L \in C(X).$$ Also,\begin{align*} \phi(U) - \phi(L) &= \phi\left(\frac{f + g + |f - g|}2\right) - \phi\left(\frac{f + g -|f - g|}2\right) \newline &= M(|f-g|) \newline &\geq \left|M(f-g)\right| \newline &= \left|\phi(f) - \phi(g)\right| \newline &\geq (1 + \varepsilon) d_\infty(f, g) = (1 + \varepsilon) d_\infty(U, L). \end{align*} Since $$U > L,$$ $$tU + (1-t)L$$ is an increasing function of $$t.$$ Take $t < \dfrac{\psi(L) - \phi(L)}{\varepsilon \cdot d_\infty(U,L)}.$ If the right-hand side is positive, $$\phi(L) < \psi(L),$$ which is a contradiction. So, $$t < 0,$$ and we have \begin{align*} \phi(tU + (1-t)L) &= t\phi(U) + (1-t)\phi(L) \newline &= \phi(L) + t\cdot\left(\phi(U) - \phi(L)\right) \newline &\leq \phi(L) + t\cdot (1 + \varepsilon) d_\infty(U, L) \newline &< \phi(L) + \left(\frac{\psi(L) - \phi(L)}{\varepsilon \cdot d_\infty(U, L)}\right)\cdot \varepsilon \cdot d_\infty(U, L) + t\cdot d_\infty(U, L) \newline &= \psi(L) + t \cdot d_\infty(U, L) \newline &= \psi(L) - |t| \cdot d_\infty(U, L) \newline &= \psi(L) - d_\infty(tU, tL) \newline &= \psi(L) - d_\infty(tU + (1 - t)L, L) \newline &\leq \psi(L) - \left(\psi(L) - \psi(tU + (1 - t)L)\right) \newline &= \psi(tU + (1-t)L), \end{align*} a contradiction.

This completes the proof.

## Problem 4b

Let $$\Phi = \{\phi_i\}_{i\in I}$$ be a collection of affine functions mapping $$C(X) \to \mathbb R$$ that are monotone, 1-Lipschitz, and 0-nonnegative. Show that $$f = \inf \Phi$$ is continuous, concave, monotone, 1-Lipschitz, and 0-nonnegative.

### Solution

• 0-nonnegativity: Suppose contrariwise that $$f(0) < 0.$$ Then there must be $$\phi_i$$ with $$\phi_i(0) = f(0)/2 < 0,$$ a contradiction.
• Monotonicity: Suppose contrariwise that there exist $$m, n \in C(X)$$ with $$m < n$$ and $$f(m) > f(n).$$ By assumption, for any $$\varepsilon > 0,$$ there is a $$\phi_i$$ with $$f(n) = \phi_i(n) + \varepsilon.$$ Then we have \begin{align*} f(m) - \phi_i(m) &> f(n) - \phi_i(m) \newline &= \phi_i(n) - \phi_i(m) + \varepsilon > 0, \end{align*} a contradiction.
• 1-Lipschitz: Suppose there are functions $$m, n \in C(X)$$ and $$\delta > 0$$ with $\frac{|f(n) - f(m)|}{d_\infty(m, n)} = 1 + \delta.$ Without loss of generality, assume $$f(n) \geq f(m).$$ Let $$0 < \varepsilon < \delta \cdot d_\infty(m, n).$$ There is a $$\phi_i$$ with $$\phi_i(m) = f(m) + \varepsilon.$$ We have \begin{align*} f(n) - \phi_i(n) &= f(m) + (1 + \delta) d_\infty(m, n) - \phi_i(n) \newline &\geq f(m) + \delta \cdot d_\infty(m, n) + |\phi_i(n) - \phi_i(m)| - \phi_i(n) \newline &\geq f(m) + \delta \cdot d_\infty(m, n) - \phi_i(m) \newline &= \delta \cdot d_\infty(m, n) - \varepsilon > 0, \end{align*} a contradiction.
• Continuity: Follows immediately.
• Concavity: Suppose contrariwise that $$f$$ is not concave. Then there exist $$g, h \in C(X)$$ and $$t \in (0, 1)$$ such that $L = t \cdot f(g) + (1 - t) f(h) > f(t \cdot g + (1 - t) h) = C.$ By assumption, there is a $$\phi \in \Phi$$ with $$\phi(t \cdot g + (1 - t)h) = (L + C)/2,$$ $$\phi(g) \geq f(g),$$ and $$\phi(h) \geq f(h).$$ But since $$\phi$$ is affine, we must have $$\phi(t \cdot g + (1 - t)h) \geq L,$$ a contradiction.

This completes the proof.

## Problem 5b

Let $$m \in \mathcal M^\pm(X).$$ Show that $$\phi : C(X) \to \mathbb R$$ defined by $$\phi(g) = \int_X g(x) ~\mathrm dm$$ is a continuous linear functional.

### Solution

Lebesgue integrals are always linear. Let $$\varepsilon > 0.$$ We must produce $$\delta > 0$$ such that whenever $$d_\infty(f, g) < \delta,$$ $\left| \int_X f ~\mathrm dm - \int_X g ~\mathrm dm \right| < \varepsilon.$ If $$m$$ is the zero measure then there is nothing to prove. Suppose $$m$$ is nonzero. Let $\delta = \frac{\varepsilon}{\int_X \mathrm dm^+ + \int_X \mathrm dm^-}.$ Then we have \begin{align*} \left| \int_X f ~\mathrm dm - \int_X g ~\mathrm dm \right| &= \left| \int_X (f - g) ~\mathrm dm \right| \newline &= \left| \int_X (f-g) ~\mathrm dm^+ - \int_X (f - g) ~\mathrm dm^- \right| \newline &\leq \left|\int_X (f - g) ~\mathrm dm^+ \right| + \left| \int_X (f - g) ~\mathrm dm^- \right| \newline &\leq \int_X |f - g| ~\mathrm dm^+ + \int_X |f - g| ~\mathrm dm^- \newline &< \int_X \delta ~\mathrm dm^+ + \int_X \delta ~\mathrm dm^- \newline &= \delta \left[\int_X \mathrm dm^+ + \int_X \mathrm dm^- \right] = \varepsilon. \end{align*}

## Problem 5c

What extra conditions on a continuous linear functional $$F : C(X) \to \mathbb R$$ are required so that it corresponds to a positive measure? What about a probability measure?

### Solution

Say $$F$$ is positive if whenever $$f \geq 0,$$ $$F(f) \geq 0.$$ By Riesz-Markov-Kakutani, positive functionals correspond to positive measures. Note that this implies monotonicity by the linearity of $$F.$$

To have a probability measure, we need $$F(1) = \int_X \mathbf 1_X ~\mathrm dm = m(X) = 1.$$ Note that this implies 1-Lipschitzness: if $$d_\infty(f, g) = \varepsilon$$ and $$F$$ is positive, then \begin{align*} | F(f) - F(g) | = | F(f - g) | \leq F\left( | f-g | \right) \leq F(\varepsilon) = \varepsilon F(1) = \varepsilon. \end{align*}

## Problem 6

Let $$\phi : C(X) \to \mathbb R$$ be affine, monotone, 1-Lipschitz, and 0-nonnegative. Show that $$\phi$$ corresponds to an affine measure $$(m, b)$$ with $$0 \leq m(X) \leq 1.$$

### Solution

We have $$\phi(f) = M(f) + \phi(0),$$ where $$M$$ is linear and monotone, hence corresponds to a positive measure $$m.$$ Since $$M$$ is 1-Lipschitz, we have $$m(X) = M(1) \in [0, 1].$$

## Fundamental Theorem of Inframeasures, forward direction

Let $$\psi : C(X) \to \mathbb R$$ be concave, monotone, 1-Lipschitz, and 0-increasing. Let $$T$$ be the set of (positive) affine measures above $$\psi$$: $T = \{(m, b): m(g) + b \geq \psi(g) \text{ for all } g \in C(X)\}.$

Prove that $$\psi(g) = \inf_{(m, b) \in T} (m(g) + b).$$

### Solution

We need to show that each $$(m, b) \in T$$ corresponds to an affine $$\phi : C(X) \to \mathbb R$$ such that $$\phi \geq \psi$$ and $$\phi$$ is 1-Lipschitz, 0-nonnegative, and monotone. Define $$\phi(g) = m(g) + b.$$ Then $$\phi$$ is affine and above $$\psi$$ by definition.

• Monotonicity: Since $$m$$ is positive, $$\phi$$ is monotone.
• 0-nonnegativity: $$\phi(\mathbf 0) = m(\mathbf 0) + b \geq \phi(\mathbf 0) > 0.$$
• 1-Lipschitz: It suffices to show that $$m(\mathbf 1) \leq 1.$$ Suppose contrariwise that $$m(\mathbf 1) = 1 + \varepsilon.$$ Let $$a < (\psi(\mathbf 0) - b)/\varepsilon \leq 0.$$ Then \begin{align*} \phi(a\mathbf 1) &= a \cdot m(\mathbf 1) + b \newline &= a(1 + \varepsilon) + b \newline &< a + \varepsilon\left(\frac{\psi(\mathbf 0) - b}\varepsilon\right) + b \newline &= \psi(\mathbf 0) + a \newline &\leq \psi(a\mathbf 1), \end{align*} a contradiction.

This completes the proof.

## Fundamental Theorem of Inframeasures, backward direction

For all sets $$\Psi$$ of affine measures $$m$$ with $$0 \leq m(\mathbf 1) \leq 1,$$ show that the functional $$\psi : C(X) \to \mathbb R$$ defined by $\psi(g) = \inf_{(m, b) \in \Psi} (m(g) + b)$ is concave, monotone, 1-Lipschitz, and 0-nonnegative.

### Solution

$$\Psi$$ corresponds to a set $$\Phi$$ of affine, monotone, 1-Lipschitz, 0-nonnegative functionals via $$(m, b) \mapsto (g \mapsto m(g) + b).$$