Strict midpoint convexity $\Rightarrow$ strict convexity

Question

Hi everyone I have trouble with the following I think is something very simple, but I cannot figure out yet the correct approach for the strict inequality

If $f$ is continuous and $f$ is strict midpoint convex, i.e., $f[(x+y)/2]<1/2[ f(x)+f(y)]$, then $f$ is strict convex.

Sketch proof: By induction we can easily obtain $f(kx+(1-k)y)<kf(x)+(1-k)f(y)$ whenever $k\in \mathbb{Q}[m/2^n]\cap(0,1)$ [i.e., $k\in \mathbb{Q}\cap (0,1)$ and $k=m/2^n$]. Now define $h(t)=tf(x)+(1-t)f(y)-f(tx+(1-t)y)$, then $h$ is clearly continuous for all $t\in (0,1)$. Since $h(k)>0$ whenever $k\in \mathbb{Q}[m/2^n]\cap(0,1)$ we claim that $h(t)\ge0$ for all $t\in (0,1)$.

We argue by contradiction, suppose that there is some $t'\in (0,1)$ for which $h(t')<0$ so by continuity there is some interval around $t'$ such that for all $t$ in the interval we must have $h(t)<0$, but since $\mathbb{Q}[m/2^n]\cap(0,1)$ is dense in $(0,1)$ there is some $k=m/2^n$ in the interval and so $h(k)<0$, a contradiction. Hence $h(t)\ge 0$, i.e., $f[tx+(1-t)y]\le tf(x)+(1-t)f(y)$ for $t\in (0,1)$.

To conclude only we have to show $h(t)>0$ for each $t\in (0,1)$. Suppose to the contrary that there is some $t'$ such that $h(t')=0$, so $\lim _{t \to t'} h(t)=0$. Given $\epsilon>0$, there is some $\delta>0$ such that $|t-t'|< \delta \Rightarrow |h(t)|< \epsilon$, in particular for $k \in \mathbb{Q}[m/2^n]\cap(0,1)$ in the interval, $h(k)>0$ and then

$$f[kx+(1-k)y]<kf(x)+(1-k)f(y)<f[kx+(1-k)y]+\epsilon$$

and here is where I'm not sure of how to get the contradiction because the $k$ may change with the $\epsilon$ any idea. Thanks in advance.

Answer 1

Not that way. Assume that it is not strictly convex. Then there are $x_0,y_0$ in the domain and $t\in(0,1)$ such that

$$f(tx_0+(1-t)y_0)=tf(x_0)+(1-t)f(y_0)$$.

Now use that the function is strictly mid-point convex for the intervals $[x_0,tx_0+(1-t)y_0]$ and $[tx_0+(1-t)y_0,y_0]$.

Call $x_1$ and $y_1$ to the mid points of those intervals. Notice that the chord from $(x_1,f(x_1))$ to $(y_1,f(y_1))$ is strictly below the chord from $(x_0,f(x_0))$ to $(y_0,f(y_0))$.

Now consider the mid point $P_1$ of $[x_1,y_1]$. If this mid point is $tx_0+(1-t)y_0$ you already got your contradiction. If not, call $[x_2,y_2]$ to which ever $[x_1,P]$ or $[P,y_1]$ contains $tx_0+(1-t)y_0$.

Now repeat. Take the mid point $P_2$ of $x_2,y_2$. If $P_2$ is $tx_0+(1-t)y_0$ contradiction. Call interval $[x_3,y_3]$ to which ever $[x_2,P_2]$ or $[P_2,y_2]$ contains $tx_0+(1-t)y_0$.

...

In this way we get a sequence of nested intervals $[x_i,y_i]$ such that the values of the function is strictly below the chord at $[x_1,y_1]$, which in turn is strictly below the chord at $[x_0,y_0]$. But the values of $f$ at $x_i$ and $y_i$ converge to $f(tx_0+(1-t)y_0)$. Contradiction.

Answer 2

I think this is a simpler approach. We have either $tx+(1-t)y\le \dfrac{x+y}2$ or $tx+(1-t)y\ge\dfrac{x+y}2$. The case when is $(x+y)/2$ is trivial and the other two cases are handled similarly, so only we check one. We may assume $tx+(1-t)y<\dfrac{x+y}2$.

Since $tx+(1-t)y<\dfrac{x+y}2$ we can express it as $\,ax+(1-a)\dfrac{x+y}2$ for $a\in (0,1)$, and after a little algebra we obtain $a=2t-1$. Since $\dfrac12<t<1$, clearly $a\in (0,1)$. So

\begin{align*} f(tx+(1-t)y)=f\left(ax+(1-a)\dfrac{x+y}2\right) \\ \le a f(x)+(1-a)f\left(\dfrac{x+y}2\right)\\ <a f(x)+(1-a)\dfrac{f(x)+f(y)}2\\ =\dfrac{a+1}2f(x)+\dfrac{1-a}2f(y)\\ =tf(x)+ (1-t)f(y). \end{align*}

Then $$f(tx+(1-t)y)<tf(x)+ (1-t)f(y),$$ as was to be shown.