Let $f:S rightarrow mathbb{R}$, where S is non empty convex set in $mathbb{R}^n$, then $fleft ( x right )$ is said to be convex on S if $fleft ( lambda x_1+left ( 1-lambda right )x_2 right )leq lambda fleft ( x_1 right )+left ( 1-lambda right )fleft ( x_2 right ), forall lambda in left ( 0,1 right )$.

On the other hand, Let $f:Srightarrow mathbb{R}$, where S is non empty convex set in $mathbb{R}^n$, then $fleft ( x right )$ is said to be concave on S if $fleft ( lambda x_1+left ( 1-lambda right )x_2 right )geq lambda fleft ( x_1 right )+left ( 1-lambda right )fleft ( x_2 right ), forall lambda in left ( 0, 1 right )$.

Let $f:S rightarrow mathbb{R}$ where S is non empty convex set in $mathbb{R}^n$, then $fleft ( xright )$ is said to be strictly convex on S if $fleft ( lambda x_1+left ( 1-lambda right )x_2 right )

Let $f:S rightarrow mathbb{R}$ where S is non empty convex set in $mathbb{R}^n$, then $fleft ( xright )$ is said to be strictly concave on S if $fleft ( lambda x_1+left ( 1-lambda right )x_2 right )> lambda fleft ( x_1 right )+left ( 1-lambda right )fleft ( x_2 right ), forall lambda in left ( 0, 1 right )$.

Examples

• A linear function is both convex and concave.

• $fleft ( x right )=left | x right |$ is a convex function.

• $fleft ( x right )= frac{1}{x}$ is a convex function.

Theorem

Let $f_1,f_2,…,f_k:mathbb{R}^n rightarrow mathbb{R}$ be convex functions. Consider the function $fleft ( x right )=displaystylesumlimits_{j=1}^k alpha_jf_jleft ( x right )$ where $alpha_j>0,j=1, 2, …k,$ then $fleft ( x right )$is a convex function.

Proof

Since $f_1,f_2,…f_k$ are convex functions

Therefore, $f_ileft ( lambda x_1+left ( 1-lambda right )x_2 right )leq lambda f_ileft ( x_1 right )+left ( 1-lambda right )f_ileft ( x_2 right ), forall lambda in left ( 0, 1 right )$ and $i=1, 2,….,k$

Consider the function $fleft ( x right )$.

Therefore,

$ fleft ( lambda x_1+left ( 1-lambda right )x_2 right )$

$=displaystylesumlimits_{j=1}^k alpha_jf_jleft ( lambda x_1 +1-lambda right )x_2leq displaystylesumlimits_{j=1}^kalpha_jlambda f_jleft ( x_1 right )+left ( 1-lambda right )f_jleft ( x_2 right )$

$Rightarrow fleft ( lambda x_1+left ( 1-lambda right )x_2 right )leq lambda left ( displaystylesumlimits_{j=1}^k alpha _jf_jleft ( x_1 right ) right )+left ( displaystylesumlimits_{j=1}^k alpha _jf_jleft ( x_2 right ) right )$

$Rightarrow fleft ( lambda x_1+left ( 1-lambda right )x_2 right )leq lambda fleft ( x_2 right )leq left ( 1-lambda right )fleft ( x_2 right )$

Hence, $fleft ( xright )$ is a convex function.

Theorem

Let $fleft ( xright )$ be a convex function on a convex set $Ssubset mathbb{R}^n$ then a local minima of $fleft ( xright )$ on S is a global minima.

Proof

Let $hat{x}$ be a local minima for $fleft ( x right )$ and $hat{x}$ is not global minima.

therefore, $exists hat{x} in S$ such that $fleft ( bar{x} right )

Since $hat{x}$ is a local minima, there exists neighbourhood $N_varepsilon left ( hat{x} right )$ such that $fleft ( hat{x} right )leq fleft ( x right ), forall x in N_varepsilon left ( hat{x} right )cap S$

But $fleft ( x right )$ is a convex function on S, therefore for $lambda in left ( 0, 1 right )$

we have $lambda hat{x}+left ( 1-lambda right )bar{x}leq lambda fleft ( hat{x} right )+left ( 1-lambda right )fleft ( bar{x} right )$

$Rightarrow lambda hat{x}+left ( 1-lambda right )bar{x}

$Rightarrow lambda hat{x}+left ( 1-lambda right )bar{x}

But for some $lambda

$lambda hat{x}+left ( 1-lambda right )bar{x} in N_varepsilon left ( hat{x} right )cap S$ and $fleft ( lambda hat{x}+left ( 1-lambda right )bar{x} right )

which is a contradiction.

Hence, $bar{x}$ is a global minima.

Epigraph

let S be a non-empty subset in $mathbb{R}^n$ and let $f:S rightarrow mathbb{R}$ then the epigraph of f denoted by epi(f) or $E_f$ is a subset of $mathbb{R}^n+1$ defined by $E_f=left { left ( x,alpha right ):x in mathbb{R}^n, alpha in mathbb{R}, fleft ( x right )leq alpha right }$

Hypograph

let S be a non-empty subset in $mathbb{R}^n$ and let $f:S rightarrow mathbb{R}$, then the hypograph of f denoted by hyp(f) or $H_f=left { left ( x, alpha right ):x in mathbb{R}^n, alpha in mathbb{R}^n, alpha in mathbb{R}, fleft ( x right )geq alpha right }$

Theorem

Let S be a non-empty convex set in $mathbb{R}^n$ and let $f:S rightarrow mathbb{R}^n$, then f is convex if and only if its epigraph $E_f$ is a convex set.

Proof

Let f is a convex function.

To show $E_f$ is a convex set.

Let $left ( x_1, alpha_1 right ),left ( x_2, alpha_2 right ) in E_f,lambda inleft ( 0, 1 right )$

To show $lambda left ( x_1,alpha_1 right )+left ( 1-lambda right )left ( x_2, alpha_2 right ) in E_f$

$Rightarrow left [ lambda x_1+left ( 1-lambda right )x_2, lambda alpha_1+left ( 1-lambda right )alpha_2 right ]in E_f$

$fleft ( x_1 right )leq alpha _1, fleft ( x_2right )leq alpha _2$

Therefore, $fleft (lambda x_1+left ( 1-lambda right )x_2 right )leq lambda fleft ( x_1 right )+left ( 1-lambda right )f left ( x_2 right )$

$Rightarrow fleft ( lambda x_1+left ( 1-lambda right )x_2 right )leq lambda alpha_1+left ( 1-lambda right )alpha_2$

Converse

Let $E_f$ is a convex set.

To show f is convex.

i.e., to show if $x_1, x_2 in S,lambda left ( 0, 1right )$

$fleft ( lambda x_1+left ( 1-lambda right )x_2 right )leq lambda fleft ( x_1 right )+left ( 1-lambda right )fleft ( x_2 right )$

Let $x_1,x_2 in S, lambda in left ( 0, 1 right ),fleft ( x_1 right ), fleft ( x_2 right ) in mathbb{R}$

Since $E_f$ is a convex set, $left ( lambda x_1+left ( 1-lambda right )x_2, lambda fleft ( x_1 right )+left ( 1-lambda right )right )fleft ( x_2 right )in E_f$

Therefore, $fleft ( lambda x_1+left ( 1-lambda right )x_2 right )leq lambda fleft ( x_1 right )+left ( 1-lambda right )fleft ( x_2 right )$