View source for Convex Analysis/Strong convexity

{{definition|strongly convex function|Let <math>X</math> be a Banach space over <math>\mathbb K = \mathbb R, \mathbb C</math>. A function <math>f: X \to (- \infty, \infty]</math> is called '''strongly convex''' with parameter <math>\Lambda > 0</math> iff the following equation holds for all <math>x, y \in X</math> and <math>t \in [0,1]</math>:
:<math>f\left( tx + (1 - t)y \right) \le t f(x) + (1 - t) f(y) - \frac{t(1-t)}{2} \Lambda \|x - y\|</math>}}

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

{{proof|Since <math>\forall x \in X: f(x) \ge b</math>, the value <math>\mu: \inf_{x \in X} f(x) \in \mathbb R</math> exists. Choose a sequence <math>(x_n)_{n \in \mathbb N}</math> in <math>X</math> such that
:<math>f(x_n) \le \mu + \frac{1}{n}</math>.<ref group="Note">If <math>X</math> is separable, so that arbitrary products of nonempty open sets are nonempty, the continuity of <math>f</math> implies that the axiom of choice is not required for this construction.</ref>
<math>(x_n)_{n \in \mathbb N}</math> is a Cauchy sequence because if <math>z \in X</math> is such that <math>f(z) \le \mu + \frac{1}{n}</math> and <math>w \in X</math> is such that <math>f(w) \le \mu + \frac{1}{m}</math>, then
:<math>\mu \le f\left( \frac{z + w}{2} \right) \le \frac{1}{2} \left( f(z) + f(w) \right) - \frac{1}{8} \Lambda \|z - w\| \le \mu + \frac{1}{2} \left( \frac{1}{n} + \frac{1}{m} \right) - \frac{1}{8} \Lambda \|z - w\|</math>,
whence
:<math>\|z - w\| \le \frac{4}{\Lambda} \left( \frac{1}{n} + \frac{1}{m} \right)</math>;
in particular, if we show that a minimizer <math>x_\infty</math> exists, then it will be unique, for if we set <math>z = x_\infty</math> and call any other minimizer <math>w</math>, the above estimate holds for <math>m,n</math> arbitrary.
Since <math>X</math> is Banach, <math>(x_n)_{n \in \mathbb N}</math> is convergent, say to <math>x_\infty \in X</math>. If we show that
:<math>f(x_\infty) < \mu + \epsilon</math>
for all <math>\epsilon > 0</math>, then <math>f(x_\infty) = \mu</math>. By the continuity of <math>f</math>, choose <math>\delta > 0</math> such that <math>\|x_\infty - y\| < \delta</math> implies <math>|f(x_\infty) - f(y)| < \epsilon/2</math>. By convergence of <math>(x_n)_{n \in \mathbb N}</math> pick <math>N \in \mathbb N</math> sufficiently large so that for all <math>m \ge N</math> <math>\|x_\infty - x_m\| < \delta</math>. Then choose <math>k \ge N</math> such that <math>\frac{1}{k} \le \frac{\epsilon}{2}</math>. Then the triangle inequality implies
:<math>f(x_\infty) - \mu = |f(x_\infty) - \mu| \le |f(x_\infty) - f(x_k)| + |f(x_k) - \mu| < \frac{\epsilon}{2} + \frac{\epsilon}{2} = \epsilon</math>.}}




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