15.2 Convex function

Convex functions enjoy a lot of interesting properties, this one below is just a small list.

... equivalent definitions

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[180]Let \(C⊂ ℝ^ n\) be a convex set. Let \(f : C →ℝ\) be convex; let \(x_ 1,\ldots ,x_ n ∈ C\) and \(t_ 1,\ldots ,t_ n ∈ [0, 1]\) be such that \(∑_{i=1}^ n t_ i = 1\). Show that

\[ ∑_{i=1}^ n t_ i x_ i∈ C \]

and

\[ f \left(∑_{i=1}^ n t_ i x_ i\right)≤ ∑_{i=1}^ n t_ i f (x_ i ) ~ . \]

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[181] Let \(C⊂ ℝ^ n\) be a convex set. Let \(f: C→ ℝ\), show that \(f\) is convex if and only if the epigraph

\[ \{ (x,y)~ |~ x∈ C~ ,~ f(x)≤ y\} \]

is a convex subset of \(C× ℝ\).

Properties

The following is a list of properties for convex functions \(f:C→ℝ\) with \(C⊆ ℝ^ n\). Obviously these properties also apply when \(n=1\); but when \(n=1\) proofs are usually easier, see the next section.

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[182]Let \(C⊆ ℝ^ n\) be a convex set, and \(f:C→ℝ\) a convex function. Given \(l\in {\mathbb {R}}\), define the sublevel set as

\[ L_ l = \{ x\in ℝ^ n: f(x)\le l\} \quad . \]

Show that \(L_ l\) is a convex (possibly empty) set. Deduce that the minimum points of \(f\) are a convex (possibly empty) set. Show that if \(f\) is strictly convex there can be at most one minimum point.

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[183] Let \(C⊂ ℝ^ n\) be a convex set; suppose that \(f_ i:C →ℝ\) are convex, where \(i∈ I\) (a non-empty, and arbitrary, family of indices), and we define \(f(x)=\sup _{i∈ I} f_ i(x)\), where we suppose (for simplicity) that \(f(x){\lt}∞\) for every \(i\): show that \(f\) is convex.

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[184] Prerequisites:2,7.Difficulty:*.Let \(C⊆ ℝ^ n\) be a convex set, let \(f:C→ℝ\) be a convex function, we fix \(z∈ {{C}^\circ }\): show that there exists \(v∈ℝ^ n\) such that

\begin{equation} ∀ x∈ C, f(x)≥ f(z) + ⟨ v,x-z⟩~ ~ . \label{eq:piano_ sotto_ convessa} \end{equation}

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The plane thus defined is called support plan for \(f\) in \(z\). Note:It is preferable not to assume that \(f\) is continuous, while proving this result, as this result is generally used to prove that \(f\) is continuous!.Hidden solution: [UNACCESSIBLE UUID ’185’]

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[186] Prerequisites:327,325,3.Difficulty:*.

Let \(C⊂ ℝ^ n\) be an open convex set, and \(f:C→ℝ\) a convex function, show that \(f\) is continuous.
Note:In the case of dimension \(n=1\), the proof is much easier, see 1.

Hidden solution: [UNACCESSIBLE UUID ’187’]

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[188] Topics:subdifferential.Prerequisites:3.Difficulty:*.

Let \(C⊆ ℝ^ n\) be an open convex set, and \(f:C→ℝ\) a convex function; Given \(z∈ C\), we define the subdifferential \(∂ f(z)\) as the set of \(v\) for which the relation ?? is valid (i.e., \(∂ f(z)\) contains all vectors \(v\) defining the support planes to \(f\) in \(z\)).

\(∂ f(z)\) enjoys interesting properties.

• \(∂ f(z)\) is locally bounded: if \(z∈ C\) and \(r{\gt}0\) is such that \(B(z,2r)⊂ C\), then \(L{\gt}0\) exists such that \(∀ y∈ B(z,r)\), \(∀ v∈ ∂ f(x)\) you have \(|v|≤ L\). In particular, for every \(z∈ C\), we have that \(∂ f(z)\) is a bounded set.

• Show that \(∂ f\) is upper continuous in this sense: if \(z∈ C\) and \((z_ n)_ n⊂ C\) and \(v_ n∈∂ f(z_ n)\) and if \(z_ n→_ n z\) and \(v_ n→_ n v\) then \(v∈∂ f(z)\). In particular, for every \(z∈ C\), \(∂ f(z)\) is a closed set.

Hidden solution: [UNACCESSIBLE UUID ’189’]

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[18B] Topics:minimum. Prerequisites:5.Let \(C⊆ ℝ^ n\) be a convex set, and \(f:C→ℝ\) a convex function. Show that \(z∈ {{C}^\circ }\) is a minimum if and only if \(0∈ ∂ f(z)\).

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[18C] Prerequisites:3,5.Note:A vice versa of 2.

Let \(C⊂ ℝ^ n\) be an open convex set; suppose that \(f:C→ℝ\) is convex; sequences \((a_ h)_ h⊆ℝ,(v_ h)_ h∈ ℝ^ n\) exist (for \(h∈ℕ\)), such that \(f(x)=\sup _{h∈ℕ} (a_ h+v_ h⋅ x)\). Hidden solution: [UNACCESSIBLE UUID ’18D’]