# Convex Analysis/Strong convexity

Definition (strongly convex function):

Let ${\displaystyle X}$ be a Banach space over ${\displaystyle \mathbb {K} =\mathbb {R} ,\mathbb {C} }$. A function ${\displaystyle f:X\to (-\infty ,\infty ]}$ is called strongly convex with parameter ${\displaystyle \Lambda >0}$ iff the following equation holds for all ${\displaystyle x,y\in X}$ and ${\displaystyle t\in [0,1]}$:

${\displaystyle f\left(tx+(1-t)y\right)\leq tf(x)+(1-t)f(y)-{\frac {t(1-t)}{2}}\Lambda \|x-y\|}$

Proposition (existence and uniqueness of minimizers of strongly convex functions):

Let ${\displaystyle X}$ be a Banach space over ${\displaystyle \mathbb {K} =\mathbb {R} ,\mathbb {C} }$ and let ${\displaystyle f:X\to \mathbb {R} }$ be a strongly convex function with parameter ${\displaystyle \Lambda >0}$, which additionally is bounded below (say by ${\displaystyle b\in \mathbb {R} }$) and continuous. Then ${\displaystyle f}$ admits a unique minimizer (ie. an element which realizes the infimum of ${\displaystyle f(x)}$, where ${\displaystyle x}$ ranges over ${\displaystyle X}$).

Proof: Since ${\displaystyle \forall x\in X:f(x)\geq b}$, the value ${\displaystyle \mu :\inf _{x\in X}f(x)\in \mathbb {R} }$ exists. Choose a sequence ${\displaystyle (x_{n})_{n\in \mathbb {N} }}$ in ${\displaystyle X}$ such that

${\displaystyle f(x_{n})\leq \mu +{\frac {1}{n}}}$.[Note 1]

${\displaystyle (x_{n})_{n\in \mathbb {N} }}$ is a Cauchy sequence because if ${\displaystyle z\in X}$ is such that ${\displaystyle f(z)\leq \mu +{\frac {1}{n}}}$ and ${\displaystyle w\in X}$ is such that ${\displaystyle f(w)\leq \mu +{\frac {1}{m}}}$, then

${\displaystyle \mu \leq f\left({\frac {z+w}{2}}\right)\leq {\frac {1}{2}}\left(f(z)+f(w)\right)-{\frac {1}{8}}\Lambda \|z-w\|\leq \mu +{\frac {1}{2}}\left({\frac {1}{n}}+{\frac {1}{m}}\right)-{\frac {1}{8}}\Lambda \|z-w\|}$,

whence

${\displaystyle \|z-w\|\leq {\frac {4}{\Lambda }}\left({\frac {1}{n}}+{\frac {1}{m}}\right)}$;

in particular, if we show that a minimizer ${\displaystyle x_{\infty }}$ exists, then it will be unique, for if we set ${\displaystyle z=x_{\infty }}$ and call any other minimizer ${\displaystyle w}$, the above estimate holds for ${\displaystyle m,n}$ arbitrary. Since ${\displaystyle X}$ is Banach, ${\displaystyle (x_{n})_{n\in \mathbb {N} }}$ is convergent, say to ${\displaystyle x_{\infty }\in X}$. If we show that

${\displaystyle f(x_{\infty })<\mu +\epsilon }$

for all ${\displaystyle \epsilon >0}$, then ${\displaystyle f(x_{\infty })=\mu }$. By the continuity of ${\displaystyle f}$, choose ${\displaystyle \delta >0}$ such that ${\displaystyle \|x_{\infty }-y\|<\delta }$ implies ${\displaystyle |f(x_{\infty })-f(y)|<\epsilon /2}$. By convergence of ${\displaystyle (x_{n})_{n\in \mathbb {N} }}$ pick ${\displaystyle N\in \mathbb {N} }$ sufficiently large so that for all ${\displaystyle m\geq N}$ ${\displaystyle \|x_{\infty }-x_{m}\|<\delta }$. Then choose ${\displaystyle k\geq N}$ such that ${\displaystyle {\frac {1}{k}}\leq {\frac {\epsilon }{2}}}$. Then the triangle inequality implies

${\displaystyle f(x_{\infty })-\mu =|f(x_{\infty })-\mu |\leq |f(x_{\infty })-f(x_{k})|+|f(x_{k})-\mu |<{\frac {\epsilon }{2}}+{\frac {\epsilon }{2}}=\epsilon }$. ${\displaystyle \Box }$

1. If ${\displaystyle X}$ is separable, so that arbitrary products of nonempty open sets are nonempty, the continuity of ${\displaystyle f}$ implies that the axiom of choice is not required for this construction.