Convex Analysis/Strong convexity

Definition (strongly convex function):

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

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

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

f(x_{n})\leq \mu +{\frac {1}{n}}

.^{[Note 1]}

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

\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

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

;

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

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

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

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

.

\Box

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

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

[Note 1]