Exercise 4.24

Question

Exercise 24: Assume that $f$ is a continuous real function defined in $(a,b)$ such that $$f\biggl(\frac{x+y}{2}\biggr)\le\frac{f(x)+f(y)}{2}$$ for all $x,y\in(a,b)$. Prove
    that $f$ is convex.

analambanomenos · Accepted Answer

Fix $x,y\in(a,b)$. We can show by induction that $f(\lambda x+(1-\lambda)y)\le\lambda f(x)+(1-\lambda)f(y)$ for all rational $\lambda$ between 0 and 1 with denominator
    a power of 2. This is true for $\lambda=1/2$ by assumption. Assume that the inequality holds for all $\lambda=j/2^m$ for $m<n$ and $1<j<2^m$, and let $\lambda=j/2^n$.
    If $j$ is even, then the inequality holds by the induction hypothesis, so assume $j=2k+1$ is odd. Let $$x_1=\biggl(\frac{k}{2^{n-1}}\biggr)x+
    \biggl(1-\frac{k}{2^{n-1}}\biggr)y\quad\hbox{and}\quad y_1=\biggl(\frac{k+1}{2^{n-1}}\biggr)x+\biggl(1-\frac{k+1}{2^{n-1}}\biggr)y.$$
    Then $\lambda x+(1-\lambda)y=(x_1+y_1)/2$, so we have
    \begin{align*}
        f(\lambda x+(1-\lambda)y) &\le \frac{f(x_1)+f(y_1)}{2} \
        &\le \frac{1}{2}\biggl(\biggl(\frac{k}{2^{n-1}}\biggr)f(x)+\biggl(1-\frac{k}{2^{n-1}}\biggr)f(y)\biggr) +
            \frac{1}{2}\biggl(\biggl(\frac{k+1}{2^{n-1}}\biggr)f(x)+\biggl(1-\frac{k+1}{2^{n-1}}\biggr)f(y)\biggr) \
        &= \lambda f(x)+(1-\lambda)f(y)
    \end{align*}
    Both sides of the inequality $f(\lambda x+(1-\lambda)y)\le\lambda f(x)+(1-\lambda)f(y)$ are continuous for $\lambda\in(0,1)$, and the inequality holds for a dense subset
    of $(0,1)$. Hence it holds for all $\lambda\in(0,1)$.

Amit Awasthi · Answer

\begin{proof} Suppose that $f$ is not convex. Then, for $a < x < b, a < y < b, 0 < \lambda < 1$, $f(f(\lambda x + (1-\lambda)y) > \lambda f(x) + (1-\lambda)f(y))$ by the definition of convex. Let $z = \lambda x + (1-\lambda)y$. Assume, without a loss of generality, that $x < y$. Then construct $A=\{p\in[x,z):f(p) = \lambda x + (1-\lambda)y\}$. By construction, $A$ is bounded by $x$ below and $z$ above, and $f(x) \le \lambda x + (1-\lambda)y < f(z)$ implies that $A$ is not empty from the Intermediate Value Theorem since $f$ is continuous. The boundedness of $A$ means that we can let $\alpha = \mathrm{sup} A$. Now construct $B=\{p\in(z,y]:f(p) = \lambda x + (1-\lambda)y\}$. By construction, $B$ is bounded by $z$ below and $y$ above, and $f(z) < \lambda x + (1-\lambda)y \le f(x)$ implies that $B$ is not empty from the Intermediate Value Theorem since $f$ is continuous. The boundedness of $B$ means that we can let $\beta = \mathrm{sup} B$. Next, by construction, we also know that $\alpha \le z \le \beta$. However, we know that both $\alpha = z$ and $\beta = z$ cannot be true because then we would have a simple discontinuity at $z$, which contradicts that $f$ is continuous. So, $\alpha < z < \beta$. By construction, we can then say that for all $c \in (\alpha,\beta)$, $f(c) > \lambda f(\alpha) + (1-\lambda)f(\beta)$. Specifically for $\lambda = 1/2$, for all $c \in (\alpha,\beta)$, $f(c) > (f(\alpha) + f(\beta))/2$. However, this is a contradiction since the values of $f$ over an interval cannot be strictly greater than the average value of $f$ over this interval since $f$ is continuous. Therefore, $f$ must be convex. \end{proof}

Exercise 4.24

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Exercise 4.24

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