## Continuity and Convexity

In this blog post, I will talk about concepts of continuity and convexity. In particular, I will illustrate one important relationship between these two properties a function may have. All of my definitions and references will be from Rudin’s Principles of Mathematical Analysis. First, some definitions follow:

Definition: Suppose $X$ and $Y$ are metric spaces, $E\subset X$, $p\in E$, and $f$ maps $E$ into $Y$. Then $f$ is said to be continuous at $p$ if for every $\epsilon>0$, there exists some $\delta>0$ such that

$\displaystyle d_Y(f(x),f(p))<\epsilon$

for every point $x$ satisfying $d(x,p)<\delta$.

Remark: In the above definition, $d_X$ and $d_Y$ denote metrics (or distance function) for metric spaces $X$ and $Y$, respectively. If function $f$ is continuous at every point $p\in E$, we say that $f$ is continuous on $E$, or simply say $f$ is continuous when the domain of definition is clear from context.

Since we are going to focus on real-valued functions that are defined on intervals, we might as well give definition of continuity for this particular case.

Definition: A function $f:(a,b)\rightarrow\mathbb{R}$ is said to be continuous, if for every $p\in (a,b)$  and $\epsilon>0$, there exists $\delta>0$ such that

$\displaystyle |f(x)-f(p)|<\epsilon$

for all points $x\in (a,b)$ satisfying $|x-p|<\delta$.

Definition: A function $f:(a,b)\rightarrow\mathbb{R}$ is said to be convex, if for every $x,y\in (a,b)$ and $\lambda\in[0,1]$

$\displaystyle f(\lambda x + (1-\lambda)y)\le \lambda f(x)+ (1-\lambda) f(y)$

Geometrically this means that for a convex function, the line connecting points $(a, f(a))$ and $(b, f(b))$ lies above the graph of $f(x)$ on $(a,b)$. Some examples of convex functions would be $f(x)=x^2$, $f(x)=e^x$, etc.

It is rather a remarkable fact that every real-valued convex function on $(a,b)$ is continuous on $(a,b)$. (I won’t show the proof of this). It turns out that there is a partial converse to this statement, and this blog post will aim to prove the following result

Proposition: If $f:(a,b)\rightarrow\mathbb{R}$ is continuous, and

$\displaystyle f\left(\frac{x+y}{2}\right)\le \frac{f(x)+f(y)}{2}$

then $f$ is convex.

There is name for a function that satisfies the above property:

Definition: A function $f:(a,b)\rightarrow\mathbb{R}$ is called midpoint convex if for all $x,y\in (a,b)$

$\displaystyle f\left(\frac{x+y}{2}\right)\le \frac{f(x)+f(y)}{2}$

In other words, mid-point convexity is just a special case when one uses $\lambda=\frac{1}{2}$ in the definition of convexity. It trivially follows that every convex function is midpoint convex. (the converse is actually not true).  So, the proposition basically states that every continuous midpoint convex function is convex. Here is a proof of proposition:

Proof: Assume, to the contrary, that the function $f$ is not convex. Then, there exists some subinterval $[c,d]\subset (a,b)$ such that graph of $f$ (restricted to $[c,d]$) is not under the chord joining $(c,f(c))$ and $(d,f(d))$. Now, define

$\displaystyle g(x)=f(x)-\frac{f(d)-f(c)}{d-c}(x-c)-f(c)$

and let $\gamma=\sup\{g(x) | x\in[c,d]\}$. Since we are assuming $f$ is not convex, we must have $\gamma > 0$. Observe that $g$ is continuous (since $f$ is continuous). It is easy to verify that $g$ is also midpoint convex. Let $\xi=\inf\{x\in[c,d]|g(x)=\gamma\}$. Since $g(c)=g(d)=0$ and $\gamma>0$, we must have $\xi\in (c,d)$. It follows that for any $h>0$ satisfying $\xi\pm h\in (c,d)$ we get

$\displaystyle g(\xi-h)

which implies that

$\displaystyle g\left(\frac{\xi-h+\xi+h}{2}\right)=g(\xi)> \frac{g(\xi-h)+g(\xi+h)}{2}$

which contradicts the fact that $g$ is midpoint convex. ∎