# Convex function

## 1 Convexity definitions

### 1.1 Jensen’s inequality

The function f(x), which is defined on the convex set S \subseteq \mathbb{R}^n, is called convex on S, if:

f(\lambda x_1 + (1 - \lambda)x_2) \le \lambda f(x_1) + (1 - \lambda)f(x_2)

for any x_1, x_2 \in S and 0 \le \lambda \le 1.
If the above inequality holds as strict inequality x_1 \neq x_2 and 0 < \lambda < 1, then the function is called strictly convex on S.

Jensen’s inequality

Let f(x) be a convex function on a convex set X \subseteq \mathbb{R}^n and let x_i \in X, 1 \leq i \leq m, be arbitrary points from X. Then

f\left( \sum_{i=1}^{m} \lambda_i x_i \right) \leq \sum_{i=1}^{m} \lambda_i f(x_i)

for any \lambda = [\lambda_1, \ldots, \lambda_m] \in \Delta_m - probability simplex.

1. First, note that the point \sum_{i=1}^{m} \lambda_i x_i as a convex combination of points from the convex set X belongs to X.

2. We will prove this by induction. For m = 1, the statement is obviously true, and for m = 2, it follows from the definition of a convex function.

3. Assume it is true for all m up to m = k, and we will prove it for m = k + 1. Let \lambda \in \Delta{k+1} and

x = \sum_{i=1}^{k+1} \lambda_i x_i = \sum_{i=1}^{k} \lambda_i x_i + \lambda_{k+1} x_{k+1}.

Assuming 0 < \lambda_{k+1} < 1, as otherwise, it reduces to previously considered cases, we have

x = \lambda_{k+1} x_{k+1} + (1 - \lambda_{k+1}) \bar{x},

where \bar{x} = \sum_{i=1}^{k} \gamma_i x_i and \gamma_i = \frac{\lambda_i}{1-\lambda_{k+1}} \geq 0, 1 \leq i \leq k.

4. Since \lambda \in \Delta_{k+1}, then \gamma = [\gamma_1, \ldots, \gamma_k] \in \Delta_k. Therefore \bar{x} \in X and by the convexity of f(x) and the induction hypothesis:

\begin{split} f\left( \sum_{i=1}^{k+1} \lambda_i x_i \right) = f\left( \lambda_{k+1} x_{k+1} + (1 - \lambda_{k+1})\bar{x} \right) &\leq \\ \lambda_{k+1}f(x_{k+1}) + (1 - \lambda_{k+1})f(\bar{x}) \leq \sum_{i=1}^{k+1} \lambda_i f(x_i)& \end{split}

Thus, initial inequality is satisfied for m = k + 1 as well.

Example
• f(x) = x^p, \; p > 1,\; x \in \mathbb{R}_+
• f(x) = \|x\|^p,\; p > 1, x \in \mathbb{R}^n
• f(x) = e^{cx},\; c \in \mathbb{R}, x \in \mathbb{R}
• f(x) = -\ln x,\; x \in \mathbb{R}_{++}
• f(x) = x\ln x,\; x \in \mathbb{R}_{++}
• The sum of the largest k coordinates f(x) = x_{(1)} + \ldots + x_{(k)},\; x \in \mathbb{R}^n
• f(X) = \lambda_{max}(X),\; X = X^T
• f(X) = - \log \det X, \; X \in S^n_{++}

### 1.2 Epigraph

For the function f(x), defined on S \subseteq \mathbb{R}^n, the following set:

\text{epi } f = \left\{[x,\mu] \in S \times \mathbb{R}: f(x) \le \mu\right\}

is called epigraph of the function f(x).

Convexity of the epigraph is the convexity of the function

For a function f(x), defined on a convex set X, to be convex on X, it is necessary and sufficient that the epigraph of f is a convex set.

1. Necessity: Assume f(x) is convex on X. Take any two arbitrary points [x_1, \mu_1] \in \text{epi}f and [x_2, \mu_2] \in \text{epi}f. Also take 0 \leq \lambda \leq 1 and denote x_{\lambda} = \lambda x_1 + (1 - \lambda) x_2, \mu_{\lambda} = \lambda \mu_1 + (1 - \lambda) \mu_2. Then,

\lambda\begin{bmatrix}x_1\\ \mu_1\end{bmatrix} + (1 - \lambda)\begin{bmatrix}x_2\\ \mu_2\end{bmatrix} = \begin{bmatrix}x_{\lambda}\\ \mu_{\lambda}\end{bmatrix}.

From the convexity of the set X, it follows that x_{\lambda} \in X. Moreover, since f(x) is a convex function,

f(x_{\lambda}) \leq \lambda f(x_1) + (1 - \lambda) f(x_2) \leq \lambda \mu_1 + (1 - \lambda) \mu_2 = \mu_{\lambda}

Inequality above indicates that \begin{bmatrix}x_{\lambda}\\ \mu_{\lambda}\end{bmatrix} \in \text{epi}f. Thus, the epigraph of f is a convex set.

2. Sufficiency: Assume the epigraph of f, \text{epi}f, is a convex set. Then, from the membership of the points [x_1, \mu_1] and [x_2, \mu_2] in the epigraph of f, it follows that

\begin{bmatrix}x_{\lambda}\\ \mu_{\lambda}\end{bmatrix} = \lambda\begin{bmatrix}x_1\\ \mu_1\end{bmatrix} + (1 - \lambda)\begin{bmatrix}x_2\\ \mu_2\end{bmatrix} \in \text{epi}f

for any 0 \leq \lambda \leq 1, i.e., f(x_{\lambda}) \leq \mu_{\lambda} = \lambda \mu_1 + (1 - \lambda) \mu_2. But this is true for all \mu_1 \geq f(x_1) and \mu_2 \geq f(x_2), particularly when \mu_1 = f(x_1) and \mu_2 = f(x_2). Hence we arrive at the inequality

f(x_{\lambda}) = f (\lambda x_1 + (1 - \lambda) x_2) \leq \lambda f(x_1) + (1 - \lambda) f(x_2).

Since points x_1 \in X and x_2 \in X can be arbitrarily chosen, f(x) is a convex function on X.

### 1.3 Sublevel set

For the function f(x), defined on S \subseteq \mathbb{R}^n, the following set:

\mathcal{L}_\beta = \left\{ x\in S : f(x) \le \beta\right\}

is called sublevel set or Lebesgue set of the function f(x).

## 2 Criteria of convexity

### 2.1 First-order differential criterion of convexity

The differentiable function f(x) defined on the convex set S \subseteq \mathbb{R}^n is convex if and only if \forall x,y \in S:

f(y) \ge f(x) + \nabla f^T(x)(y-x)

Let y = x + \Delta x, then the criterion will become more tractable:

f(x + \Delta x) \ge f(x) + \nabla f^T(x)\Delta x

### 2.2 Second-order differential criterion of convexity

Twice differentiable function f(x) defined on the convex set S \subseteq \mathbb{R}^n is convex if and only if \forall x \in \mathbf{int}(S) \neq \emptyset:

\nabla^2 f(x) \succeq 0

In other words, \forall y \in \mathbb{R}^n:

\langle y, \nabla^2f(x)y\rangle \geq 0

### 2.3 Connection with epigraph

The function is convex if and only if its epigraph is a convex set.

Example

Let a norm \Vert \cdot \Vert be defined in the space U. Consider the set:

K := \{(x,t) \in U \times \mathbb{R}^+ : \Vert x \Vert \leq t \}

which represents the epigraph of the function x \mapsto \Vert x \Vert. This set is called the cone norm. According to the statement above, the set K is convex.

In the case where U = \mathbb{R}^n and \Vert x \Vert = \Vert x \Vert_2 (Euclidean norm), the abstract set K transitions into the set:

\{(x,t) \in \mathbb{R}^n \times \mathbb{R}^+ : \Vert x \Vert_2 \leq t \}

### 2.4 Connection with sublevel set

If f(x) - is a convex function defined on the convex set S \subseteq \mathbb{R}^n, then for any \beta sublevel set \mathcal{L}_\beta is convex.

The function f(x) defined on the convex set S \subseteq \mathbb{R}^n is closed if and only if for any \beta sublevel set \mathcal{L}_\beta is closed.

### 2.5 Reduction to a line

f: S \to \mathbb{R} is convex if and only if S is a convex set and the function g(t) = f(x + tv) defined on \left\{ t \mid x + tv \in S \right\} is convex for any x \in S, v \in \mathbb{R}^n, which allows checking convexity of the scalar function to establish convexity of the vector function.

## 3 Strong convexity

f(x), defined on the convex set S \subseteq \mathbb{R}^n, is called \mu-strongly convex (strongly convex) on S, if:

f(\lambda x_1 + (1 - \lambda)x_2) \le \lambda f(x_1) + (1 - \lambda)f(x_2) - \frac{\mu}{2} \lambda (1 - \lambda)\|x_1 - x_2\|^2

for any x_1, x_2 \in S and 0 \le \lambda \le 1 for some \mu > 0.

### 3.1 Criteria of strong convexity

#### 3.1.1 First-order differential criterion of strong convexity

Differentiable f(x) defined on the convex set S \subseteq \mathbb{R}^n is \mu-strongly convex if and only if \forall x,y \in S:

f(y) \ge f(x) + \nabla f^T(x)(y-x) + \dfrac{\mu}{2}\|y-x\|^2

Let y = x + \Delta x, then the criterion will become more tractable:

f(x + \Delta x) \ge f(x) + \nabla f^T(x)\Delta x + \dfrac{\mu}{2}\|\Delta x\|^2

#### 3.1.2 Second-order differential criterion of strong convexity

Twice differentiable function f(x) defined on the convex set S \subseteq \mathbb{R}^n is called \mu-strongly convex if and only if \forall x \in \mathbf{int}(S) \neq \emptyset:

\nabla^2 f(x) \succeq \mu I

In other words:

\langle y, \nabla^2f(x)y\rangle \geq \mu \|y\|^2

Theorem

Let f(x) be a differentiable function on a convex set X \subseteq \mathbb{R}^n. Then f(x) is strongly convex on X with a constant \mu > 0 if and only if

f(x) - f(x_0) \geq \langle \nabla f(x_0), x - x_0 \rangle + \frac{\mu}{2} \| x - x_0 \|^2

for all x, x_0 \in X.

Necessity: Let 0 < \lambda \leq 1. According to the definition of a strongly convex function,

f(\lambda x + (1 - \lambda) x_0) \leq \lambda f(x) + (1 - \lambda) f(x_0) - \frac{\mu}{2} \lambda (1 - \lambda) \| x - x_0 \|^2

or equivalently,

f(x) - f(x_0) - \frac{\mu}{2} (1 - \lambda) \| x - x_0 \|^2 \geq \frac{1}{\lambda} [f(\lambda x + (1 - \lambda) x_0) - f(x_0)] =

= \frac{1}{\lambda} [f(x_0 + \lambda(x - x_0)) - f(x_0)] = \frac{1}{\lambda} [\lambda \langle \nabla f(x_0), x - x_0 \rangle + o(\lambda)] =

= \langle \nabla f(x_0), x - x_0 \rangle + \frac{o(\lambda)}{\lambda}.

Thus, taking the limit as \lambda \downarrow 0, we arrive at the initial statement.

Sufficiency: Assume the inequality in the theorem is satisfied for all x, x_0 \in X. Take x_0 = \lambda x_1 + (1 - \lambda) x_2, where x_1, x_2 \in X, 0 \leq \lambda \leq 1. According to the inequality, the following inequalities hold:

f(x_1) - f(x_0) \geq \langle \nabla f(x_0), x_1 - x_0 \rangle + \frac{\mu}{2} \| x_1 - x_0 \|^2,

f(x_2) - f(x_0) \geq \langle \nabla f(x_0), x_2 - x_0 \rangle + \frac{\mu}{2} \| x_2 - x_0 \|^2.

Multiplying the first inequality by \lambda and the second by 1 - \lambda and adding them, considering that

x_1 - x_0 = (1 - \lambda)(x_1 - x_2), \quad x_2 - x_0 = \lambda(x_2 - x_1),

and \lambda(1 - \lambda)^2 + \lambda^2(1 - \lambda) = \lambda(1 - \lambda), we get

\begin{split} \lambda f(x_1) + (1 - \lambda) f(x_2) - f(x_0) - \frac{\mu}{2} \lambda (1 - \lambda) \| x_1 - x_2 \|^2 \geq \\ \langle \nabla f(x_0), \lambda x_1 + (1 - \lambda) x_2 - x_0 \rangle = 0. \end{split}

Thus, inequality from the definition of a strongly convex function is satisfied. It is important to mention, that \mu = 0 stands for the convex case and corresponding differential criterion.

Theorem

Let X \subseteq \mathbb{R}^n be a convex set, with \text{int}X \neq \emptyset. Furthermore, let f(x) be a twice continuously differentiable function on X. Then f(x) is strongly convex on X with a constant \mu > 0 if and only if

\langle y, \nabla^2 f(x) y \rangle \geq \mu \| y \|^2 \quad

for all x \in X and y \in \mathbb{R}^n.

The target inequality is trivial when y = \mathbf{0}_n, hence we assume y \neq \mathbf{0}_n.

Necessity: Assume initially that x is an interior point of X. Then x + \alpha y \in X for all y \in \mathbb{R}^n and sufficiently small \alpha. Since f(x) is twice differentiable,

f(x + \alpha y) = f(x) + \alpha \langle \nabla f(x), y \rangle + \frac{\alpha^2}{2} \langle y, \nabla^2 f(x) y \rangle + o(\alpha^2).

Based on the first order criterion of strong convexity, we have

\frac{\alpha^2}{2} \langle y, \nabla^2 f(x) y \rangle + o(\alpha^2) = f(x + \alpha y) - f(x) - \alpha \langle \nabla f(x), y \rangle \geq \frac{\mu}{2} \alpha^2 \| y \|^2.

This inequality reduces to the target inequality after dividing both sides by \alpha^2 and taking the limit as \alpha \downarrow 0.

If x \in X but x \notin \text{int}X, consider a sequence \{x_k\} such that x_k \in \text{int}X and x_k \rightarrow x as k \rightarrow \infty. Then, we arrive at the target inequality after taking the limit.

Sufficiency: Using Taylor’s formula with the Lagrange remainder and the target inequality, we obtain for x + y \in X:

f(x + y) - f(x) - \langle \nabla f(x), y \rangle = \frac{1}{2} \langle y, \nabla^2 f(x + \alpha y) y \rangle \geq \frac{\mu}{2} \| y \|^2,

where 0 \leq \alpha \leq 1. Therefore,

f(x + y) - f(x) \geq \langle \nabla f(x), y \rangle + \frac{\mu}{2} \| y \|^2.

Consequently, by the first order criterion of strong convexity, the function f(x) is strongly convex with a constant \mu. It is important to mention, that \mu = 0 stands for the convex case and corresponding differential criterion.

## 4 Facts

• f(x) is called (strictly) concave, if the function -f(x) - is (strictly) convex.

• Jensen’s inequality for the convex functions:

f \left( \sum\limits_{i=1}^n \alpha_i x_i \right) \leq \sum\limits_{i=1}^n \alpha_i f(x_i)

for \alpha_i \geq 0; \quad \sum\limits_{i=1}^n \alpha_i = 1 (probability simplex)
For the infinite dimension case:

f \left( \int\limits_{S} x p(x)dx \right) \leq \int\limits_{S} f(x)p(x)dx

If the integrals exist and p(x) \geq 0, \quad \int\limits_{S} p(x)dx = 1.

• If the function f(x) and the set S are convex, then any local minimum x^* = \text{arg}\min\limits_{x \in S} f(x) will be the global one. Strong convexity guarantees the uniqueness of the solution.

• Let f(x) - be a convex function on a convex set S \subseteq \mathbb{R}^n. Then f(x) is continuous \forall x \in \textbf{ri}(S).

## 5 Operations that preserve convexity

• Non-negative sum of the convex functions: \alpha f(x) + \beta g(x), (\alpha \geq 0 , \beta \geq 0).
• Composition with affine function f(Ax + b) is convex, if f(x) is convex.
• Pointwise maximum (supremum) of any number of functions: If f_1(x), \ldots, f_m(x) are convex, then f(x) = \max \{f_1(x), \ldots, f_m(x)\} is convex.
• If f(x,y) is convex on x for any y \in Y: g(x) = \underset{y \in Y}{\operatorname{sup}}f(x,y) is convex.
• If f(x) is convex on S, then g(x,t) = t f(x/t) - is convex with x/t \in S, t > 0.
• Let f_1: S_1 \to \mathbb{R} and f_2: S_2 \to \mathbb{R}, where \operatorname{range}(f_1) \subseteq S_2. If f_1 and f_2 are convex, and f_2 is increasing, then f_2 \circ f_1 is convex on S_1.

## 6 Other forms of convexity

• Log-convex: \log f is convex; Log convexity implies convexity.
• Log-concavity: \log f concave; not closed under addition!
• Exponentially convex: [f(x_i + x_j )] \succeq 0, for x_1, \ldots , x_n
• Operator convex: f(\lambda X + (1 − \lambda )Y ) \preceq \lambda f(X) + (1 − \lambda )f(Y)
• Quasiconvex: f(\lambda x + (1 − \lambda) y) \leq \max \{f(x), f(y)\}
• Pseudoconvex: \langle \nabla f(y), x − y \rangle \geq 0 \longrightarrow f(x) \geq f(y)
• Discrete convexity: f : \mathbb{Z}^n \to \mathbb{Z}; “convexity + matroid theory.”
Example

Show, that f(x) = c^\top x + b is convex and concave.

Example

Show, that f(x) = x^\top Ax, where A\succeq 0 - is convex on \mathbb{R}^n.

Example

Show, that f(A) = \lambda_{max}(A) - is convex, if A \in S^n.

Example

PL inequality holds if the following condition is satisfied for some $> 0 ,$ f(x) ^2 (f(x) - f^*) x The example of a function, that satisfies the PL-condition, but is not convex. f(x,y) =