Table of Contents
These are S’s solutions for the first set of infraBayesian 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!
Notes
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 orderpreserving.
I shall use the terminology 0nonnegative rather than 0increasing 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 nonexpansive for 1Lipschitzness, 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 HahnBanach 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 HahnBanach 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, 0nonnegative, and 1Lipschitz. 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, 0nonnegative, and 1Lipschitz.
 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.
 1Lipschitz: 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.
 0nonnegativity: \(\phi(0) \geq f(0) \geq 0.\)
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, 1Lipschitz, and 0nonnegative. Show that \(f(x) = \inf \Phi(x)\) is continuous, concave, monotone, 1Lipschitz, and 0nonnegative.
Solution

0nonnegativity: \(f(0) = \inf \Phi(0) = C.\) 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.

1Lipschitz: Suppose towards a contradiction that \(f\) is not 1Lipschitz. 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 \[ \begin{align*} f(y)  \phi_i(y) &\geq f(y)  \left(\phi_i(x) + y  x\right) \newline &= f(y)  \left(f(x) + \varepsilon + y  x\right) \newline &= [f(y)  f(x)]  \varepsilon  y  x \newline &= Ly  x  \varepsilon  y  x \newline &= (L  1)y  x  \varepsilon > 0, \end{align*}\] 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 \(\phi(tx + (1t)y) = (L + C)/2.\) We must have \(\phi(x) \geq f(x)\) and \(\phi(y) \geq f(y).\) But since \(\phi\) is affine, we must have \(\phi(tx + (1t)y) \geq L,\) a contradiction.
Problem 4a
Let \(\psi : C(X) \to \mathbb R\) be concave, monotone, 1Lipschitz, and 0nonnegative. Let \(\Phi\) be the set of all affine, monotone, 1Lipschitz, and 0nonnegative 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, \[ t \cdot \psi(f_1) + (1  t)\psi(f_2) \leq \psi(t\cdot f_1 + (1t) f_2). \]
Therefore HahnBanach 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 HahnBanach 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 HahnBanach 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 HahnBanach condition to \((f, \psi(f))\) gives \(\psi(f) \leq \phi(f) < k,\) as desired.
It remains to show that \(\phi\) is monotone, 1Lipschitz, and 0nonnegative.
 0nonnegativity: \(\phi(0) \geq \psi(0) \geq 0.\)
 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.
 1Lipschitz: 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,\[\begin{align*} \phi(U)  \phi(L) &= \phi\left(\frac{f + g + f  g}2\right)  \phi\left(\frac{f + g f  g}2\right) \newline &= M(fg) \newline &\geq \leftM(fg)\right \newline &= \left\phi(f)  \phi(g)\right \newline &\geq (1 + \varepsilon) d_\infty(f, g) = (1 + \varepsilon) d_\infty(U, L). \end{align*}\] Since \(U > L,\) \(tU + (1t)L\) is an increasing function of \(t.\) Take \[t < \dfrac{\psi(L)  \phi(L)}{\varepsilon \cdot d_\infty(U,L)}.\] If the righthand side is positive, \(\phi(L) < \psi(L),\) which is a contradiction. So, \(t < 0,\) and we have \[\begin{align*} \phi(tU + (1t)L) &= t\phi(U) + (1t)\phi(L) \newline &= \phi(L) + t\cdot\left(\phi(U)  \phi(L)\right) \newline &\leq \phi(L) + t\cdot (1 + \varepsilon) d_\infty(U, L) \newline &< \phi(L) + \left(\frac{\psi(L)  \phi(L)}{\varepsilon \cdot d_\infty(U, L)}\right)\cdot \varepsilon \cdot d_\infty(U, L) + t\cdot d_\infty(U, L) \newline &= \psi(L) + t \cdot d_\infty(U, L) \newline &= \psi(L)  t \cdot d_\infty(U, L) \newline &= \psi(L)  d_\infty(tU, tL) \newline &= \psi(L)  d_\infty(tU + (1  t)L, L) \newline &\leq \psi(L)  \left(\psi(L)  \psi(tU + (1  t)L)\right) \newline &= \psi(tU + (1t)L), \end{align*}\] 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, 1Lipschitz, and 0nonnegative. Show that \(f = \inf \Phi\) is continuous, concave, monotone, 1Lipschitz, and 0nonnegative.
Solution
 0nonnegativity: 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.
 1Lipschitz: 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 \[\begin{align*} \left \int_X f ~\mathrm dm  \int_X g ~\mathrm dm \right &= \left \int_X (f  g) ~\mathrm dm \right \newline &= \left \int_X (fg) ~\mathrm dm^+  \int_X (f  g) ~\mathrm dm^ \right \newline &\leq \left\int_X (f  g) ~\mathrm dm^+ \right + \left \int_X (f  g) ~\mathrm dm^ \right \newline &\leq \int_X f  g ~\mathrm dm^+ + \int_X f  g ~\mathrm dm^ \newline &< \int_X \delta ~\mathrm dm^+ + \int_X \delta ~\mathrm dm^ \newline &= \delta \left[\int_X \mathrm dm^+ + \int_X \mathrm dm^ \right] = \varepsilon. \end{align*}\]
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 RieszMarkovKakutani, 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 1Lipschitzness: if \(d_\infty(f, g) = \varepsilon\) and \(F\) is positive, then \[\begin{align*}  F(f)  F(g)  =  F(f  g)  \leq F\left(  fg  \right) \leq F(\varepsilon) = \varepsilon F(1) = \varepsilon. \end{align*}\]
Problem 6
Let \(\phi : C(X) \to \mathbb R\) be affine, monotone, 1Lipschitz, and 0nonnegative. 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 1Lipschitz, 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, 1Lipschitz, and 0increasing. 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 1Lipschitz, 0nonnegative, 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.
 0nonnegativity: \(\phi(\mathbf 0) = m(\mathbf 0) + b \geq \phi(\mathbf 0) > 0.\)
 1Lipschitz: 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, 1Lipschitz, and 0nonnegative.
Solution
\(\Psi\) corresponds to a set \(\Phi\) of affine, monotone, 1Lipschitz, 0nonnegative functionals via \((m, b) \mapsto (g \mapsto m(g) + b).\)