We discuss a few special properties of one dimensional function convex function. The first asserts about the slope of one dimensional convex function.

Lemma 1Let be a (strict) convex function, then the function is non-decreasing (strictly increasing).

*Proof:* By the definition of convex function, we have for any

Rearrange the above inequality gives

The case for strict convexity is simply replacing the above by .

The second asserts that slope should always increases.

Lemma 2If where is any connected set in ,i.e., an interval which can be half open half closed or open or closed. The for any , we have

*Proof:* This is just a twist of algebra.

where the last one holds due to the definition of convexity.

The next lemma shows that convex function in a real line must be always increasing or always decreasing or first decreasing and then increasing.

Lemma 3Let where is an interval in . The interval can be any one kind of where . Then is either always increasing or always decreasing or first decreasing and then increasing.

*Proof:* We may suppose is not always a constant since in this case, the assertion is true. Now suppose that is not always increasing and is also not always decreasing. Then there exists all in such that

or

The latter case is not possible because it implies that

which contradicts the convexity of . Thus only the first case is possible.

Now if are interior point of , Then is continuous on the interior of and so is continuous on . This means that attains a minimum on at some as . Now for any , by convexity and monotonicity of slope, we have

which implies . Thus is decreasing on . Similarly, using monotonicity of slope, we have is increasing on . Now if suppose either or is the end point of . We take the half points . Then there are only three possible cases

The second case is ideal as and are interior point and we can proceed our previous argument and show is first decreasing and the then increasing. In the first case, we can take and ask the relation between and . The only non-ideal case is that (the other case gives three interior point such that and we can employ our previous argument). We can then further have . Again there will be only one non-ideal case that .

Continuing this process, we either stop at some point to obtain a pattern or the sequence satisfies for all and . If is an interior point, then continuity of implies that . But this is not possible because . Thus must be the end point of . Moreover, is actually decreasing on . Indeed, for any for , there is a such that and so

But since , we see is indeed decreasing on and increase/jump at the point .

The third case is similar, we will either have always increasing and only jump down at the end point or is just first decreases for some interval and then increases for some interval.

Utilizing the above lemma, we can say 1) one dimensional convex function is almost everywhere differentiable and 2) something about the boundary points of on a finite interval.

Lemma 4Suppose is convex for some . Then .

The above lemma shows that a convex function on a finite open interval is uniformly continuous as it can be extend to the boundary. However, this property does not hold in higher dimension by considering on and the point .