Exercise 4.23

Question

Exercise 23:
    A real-valued function $f$ defined in $(a,b)$ is said to be {\it convex} if $$f\bigl(\lambda x+(1-\lambda x)y\bigr)\le \lambda f(x)+(1-\lambda)f(y)$$ whenever $a<x<b$, $a<y<b$,
    $0<\lambda<1$. Prove that every convex function is continuous. Prove that every increasing convex function of a convex function is convex. If $f$ is convex in $(a,b)$, and if
    $a<s<t<u<b$, show that $$\frac{f(t)-f(s)}{t-s}\le\frac{f(u)-f(s)}{u-s}\le\frac{f(u)-f(t)}{u-t}.$$

analambanomenos · Accepted Answer

Letting $a<s<t<u<b$, then $t=\lambda u+(1+\lambda)s$ for $\lambda=(t-s)/(u-s)$. Hence
    \begin{align*}
        f(t) &\le \lambda f(u)+(1-\lambda)f(s) \
        f(t)-f(s) &\le \lambda\bigl(f(u)-f(s)\bigr) \
        \frac{f(t)-f(s)}{t-s} &\le \frac{f(u)-f(s)}{u-s}
    \end{align*}
    Also, $t=\lambda' s+(1-\lambda')u$ for $\lambda' =(u-t)/(u-s)$. Hence
    \begin{align*}
        f(t) &\le \lambda' f(s)+(1-\lambda')f(u) \
        \lambda'\bigl(f(u)-f(s)\bigr) &\le f(u)-f(t) \
        \frac{f(u)-f(s)}{u-s} &\le \frac{f(u)-f(t)}{u-t}
    \end{align*}

The first inequality can be rewritten $f(t)-f(s)=K(t-s)$ for $s<t<u$, which shows that $f$ is continuous from the right at $s$. Similarly, the second inequality can be
    rewritten $f(u)-f(t)=K'(u-t)$ for $s<t<u$, which shows that $f$ is continuous from the left at $u$. Since $s$ and $u$ were arbitrary elements of $(a,b)$, we see that $f$
    is continuous on $(a,b)$.

Let $f$ be a convex function defined in $(a,b)$ and let $g$ be an increasing convex function defined on the range of $f$. Then for  $a<x<b$, $a<y<b$, $0<\lambda<0$, we have
    \begin{align*}
        g\Bigl(f\bigl(\lambda x+(1-\lambda)y\bigr)\Bigr) &\le g\bigl(\lambda f(x)+(1-\lambda)f(y)\bigr) \
        &\le \lambda g\bigl(f(x)\bigr)+(1-\lambda)g\bigl(f(y)\bigr).
    \end{align*}
    Hence $g\circ f$ is convex on $(a,b)$.

Amit Awasthi · Answer

\begin{proof} For this function $f$, \begin{multline*} (x-z)f(y) = (z-y)f(y) + (y-x)f(y)\ > (z-y)f(x) + (y-x)f(x) \end{multline*} so \begin{equation*} f(y) > \frac{z-y}{z-x}f(x) + \frac{y-x}{z-x}f(x). \end{equation*} If $\lambda = (z-y)/(z-x)$, it is evidently in the interval $(0,1)$. Also, $\lambda x + (1 - \lambda)z = y$. This means \begin{equation*} f(\lambda x + (1-\lambda)z) > \lambda f(x) + (1-\lambda)f(z), \end{equation*} which contradicts that $f$ is convex. \end{proof} \begin{lemma} For a convex function $f$ over interval $(a,b)$, the function is monotonic for some $\epsilon > 0$ on the interval $(x,x+\epsilon)$ for any $x \in (a,b)$. \end{lemma} \begin{proof} Suppose not, and let the point $x$ be the point that makes the value of $f$ over this interval not monotone. From the previous proof, there must be some points $a,b,c \in (x,x+\epsilon)$ such that \begin{gather*} f(a) \le f(b) < f(c) \end{gather*} or \begin{gather*} f(a) < f(b) \le f(c). \end{gather*} Put $0 < \epsilon' < a-x$, there must exist $a',b' \in (x,x+\epsilon')$ such that \begin{align*} f(a') \le f(b')\ & ext{and}\ f(c') < f(b'),\ \end{align*} or \begin{align*} f(a') < f(b')\ & ext{and}\ f(c') \le f(b'), \end{align*} so that $f$ is not monotone in $(x,x+\epsilon')$. However, this gives for $b' < a < b$, \begin{align*} f(b') \le f(a)\ ext{and}\ f(a) > f(b), \end{align*} or \begin{align*} f(b') < f(a)\ ext{and}\ f(a) \ge f(b), \end{align*} which forms a contradiction. Thus, $f$ must be monotonic on $(x,x+\epsilon)$. \end{proof} \begin{lemma} A convex function $f$ over a monotone interval $(a,b)$ cannot have any simple discontinuities, and thus, cannot have any discontinuities. \end{lemma} \begin{proof} Assume that $z$ produces a simple discontinuity at $f(z)$. First $f(z)$ must be defined since otherwise \begin{equation*} f(\lambda x + (1-\lambda)y) \le \lambda f(x) + (1-\lambda)f(y) \end{equation*} would not hold when $z = \lambda x + (1-\lambda)y$, which means $f(p)$ is defined on all $p \in (a,b)$. Since $f(z)$ is a simple discontinuity, therefore, either $f(z-) e f(z)$ or $f(z+) e f(z)$. \par Now assume, without a loss of generality, that $f(z-) > f(z)$. Then, given $\epsilon > 0$ that is arbitrarily close to $0$ we can construct $\lambda f(z-\epsilon) + (1-\lambda)f(z)$. However, for $f(\lambda x + (1-\lambda)y)$, when $z - \epsilon = \lambda x + (1-\lambda)y$ and $\epsilon ightarrow 0$, $\lambda ightarrow 0$, \begin{multline*} f(\lambda x + (1-\lambda)y) = f(z-\epsilon) = f(z-)\ > f(z) = \lambda f(z-\epsilon) + (1-\lambda)f(z), \end{multline*} which contradicts the convexity of $f$. The other cases, for $f(z-) < f(z)$, $f(z+) > f(z)$, $f(z+) < f(z)$ follow by the same argument. Thus, monotonic subintervals of a convex function $f$ must not have any simple discontinuities, and therefore, discontinuities in general since monotonic functions can only have simple discontinuities. \end{proof} \begin{proof}[Proof of $(a)$] Since the monontonic sub\-in\-ter\-vals of $f$ are continuous since $f$ itself has no discontinuities, then $f$ is continuous. \end{proof} \begin{proof}[Proof of $(b)$] Let $f(x), g(x)$ be increasing and convex. Then, for $f(x)$, \begin{equation*} f(\lambda x + (1-\lambda) y) \le \lambda f(x) + (1-\lambda)f(y). \end{equation*} Since $g(x)$ is increasing, we can say \begin{equation*} g(f(\lambda x + (1-\lambda) y)) \le g(\lambda f(x) + (1-\lambda)f(y)). \end{equation*} Also, since $g(x)$ itself is convex, we can say \begin{equation*} g(\lambda f(x) + (1-\lambda) g(y)) \le \lambda g(f(x)) + (1-\lambda)g(f(y)). \end{equation*} Therefore, \begin{equation*} g(f(\lambda x + (1-\lambda) y)) \le \lambda g(f(x)) + (1-\lambda)g(f(y)), \end{equation*} and $g(f(x))$ is convex. \end{proof} \begin{proof}[Proof of $(c)$] For the first inequality, let $\lambda = (u-t)/(u-s)$. Because of the convexity of $f$ we can say that: \begin{align*} f(t) - f(s) &\le (1-\lambda)\left(f(u)-f(s) ight)\ &\quad= \frac{t-s}{u-s} \left( f(u) - f(s) ight) \end{align*} so \begin{equation*} \frac{f(t) - f(s)}{t-s} \le \frac{f(u) - f(s)}{u-s}, \end{equation*} and we have the first inequality. Then, for the second one let $t = \lambda s + (1-\lambda) u$. Then, from the convexity of $f$: \begin{gather*} f(t) \le \frac{u-t}{u-s}f(s) + \left(1 - \frac{u-t}{u-s} ight) f(u)\ \frac{f(t)}{u-t} \le \frac{f(s)}{u-s} - \frac{f(u)}{u-s}+ \frac{f(u)}{u-t}\ \frac{f(u)-f(s)}{u-s} \le \frac{f(u)-f(t)}{u-t}, \end{gather*} and we have the second inequality. \end{proof}

Exercise 4.23

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Exercise 4.23

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