> [!definition] > > Let $V$ be a $\real$-[[Vector Space|vector space]], $U \subset V$ be a [[Convexity|convex]] set, and $f: U \to \real$ be any [[Function|function]], then $f$ is **convex** if > $ > f(tx + (1 - t)y) \le tf(x) + (1 - t)f(y) > $ > for all $x, y \in U$ and $t \in [0, 1]$. In other words, the set $\bracs{(x, y): y \ge f(x)}$ is [[Convexity|convex]]. > [!theorem] "Reverse Convex Inequality" > > Let $f: U \to \real$ be a convex function, then > $ > f(tx + (1 - t)y) \ge tf(x) + (1 - t)f(y) \quad \forall t \not\in [0, 1] > $ > for all $x, y \in U$ and $t \not\in [0, 1]$. > > *Proof*. By translating $U$ and $f$, assume without loss of generality that $y = 0$ and $f(0) = 0$. Suppose that there exists $t \not\in [0, 1]$ such that $f(tx) < tf(x)$, and assume without loss of generality that $t > 1$. In this case, by convexity, > $ > f(x) \le \frac{1}{t}f(tx) < f(x) > $ > which is impossible. > [!theorem] > > Let $\seqi{f}$ be a set of convex functions bounded from above, then $\sup_{i \in I}f_i$ is also convex. > > *Proof*. Let $x, y \in U$, then > $ > f_i(tx + (1 - t)y) \le tf_i(x) + (1 - t)f_i(y) > $ > Taking the supremum on both sides yields the desired result. # Difference Quotients > [!theorem] > > Let $f: (c, d) \to \real$ be a convex function. For each $x \in (c, d)$ and sufficiently small $t \ge 0$, denote > $ > F(x, t) = \frac{f(x + t) - f(x)}{t} > $ > as the difference quotient from the right. > 1. For fixed $x$, $F(x, \cdot)$ is non-decreasing in $t$. > 2. For fixed $t$, $F(\cdot, t)$ is non-decreasing in $x$. > > *Proof*. Fix $x \in (c, d)$. By shifting the function, assume without loss of generality that $x = 0$ and $f(x) = 0$. In this case, > $ > F(x, t) = \frac{f(t)}{t} > $ > By the previous part, if $s \ge t$, then > $ > f(s) \ge \frac{s}{t}f(t) \quad \frac{f(s)}{s} \ge \frac{f(t)}{t} > $ > On the other hand, fix $t \ge 0$ and still assume that $x = 0$ and $f(x) = 0$. Let $y \ge x$, then > $ > F(y, t) = \frac{f(y + t) - f(y)}{t} \quad F(x, t) = \frac{f(t)}{t} > $ > Suppose that $t \le y$, then > $ > f(y + t) \ge \frac{y + t}{y}f(y) = (y + t)F(0, y) = f(y) + tF(0, y) = f(y) + f(t) > $ > so $F(y, t) \ge F(x, t)$. > [!theorem] > > Let $f: (c, d) \to \real$ be a convex function, then for each $x \in (c, d)$, > $ > \lim_{t \downto 0}F(x, t) = \frac{f(x + t) - f(x)}{t} > $ > exists. > > *Proof*. Let $s \in \real$ and define > $ > L(s)= f(x) + sF(x, t) > $ > since $L(0) = f(x)$ and $L(1) = f(x + t)$, using the reverse inequality gives that > $ > f(x + st) \ge L(s) = f(x) + sF(x, t) \quad \forall s \not\in [0, 1] > $ > Let $s < 0$ such that $x + st \in (c, x)$, then > $ > \begin{align*} > f(x + st) &\ge f(x) + sF(x, t) \\ > \frac{f(x + st) - f(x)}{s} &\le F(x, t) > \end{align*} > $ > Since $s$ is fixed, and $f$ is continuous, there exists a lower bound on $F(x, t)$ for sufficiently small $t$. As $F(x, t)$ is non-increasing as $t \downto 0$, the limit exists. > [!theorem] > > Let $f: (c, d) \to \real$ and $[a, b] \subset (c, d)$. Let $0 < \delta < b - d$, then the difference quotient on $[a, b]$ with the restriction > $ > F: [a, b] \times (0, \delta) \to \real \quad (x, t) \mapsto \frac{f(x + t) - f(x)}{t} > $ > is bounded. > > *Proof*. By monotonicity, > $ > \lim_{s \downto 0}F(a, s) \le F(x, t) \le F(b, \delta) > $