Why do the coefficients have to sum up to 1 in a convex function?

I was studying convex functions for convex optimization and ran into a question I'm having difficulty finding the answer to.

I noticed that the definition for a convex function is as follows:

$$\forall{x_1, x_2} \in X,\ \forall{t} \in [0,\ 1]:\quad f(tx_1 + (1-t)x_2) \le tf(x_1) + (1 - t)f(x_2)$$

This definition is from Wikipedia, but I also noticed in my textbook (Convex Optimization (Boyd & Vandenberghe)) they use $$\alpha$$ and $$\beta$$ for the coefficients, but also make sure to specify that $$\alpha + \beta = 1$$.

This question is probably due to me lacking something relatively elementary, but why must they sum up to $$1$$?

• This is nothing magical. That’s just a standard parameterization of a line segment. Mar 8, 2019 at 4:52
• The affine combination of two points is the line connecting them. The convex combination of two points is the line segment connecting them. If you're curious about convex combinations, start with affine combinations. Mar 9, 2019 at 16:11

This definition captures the idea that the graph of a convex function is always below the secant joining any two points.

Given two points $$u_1 = (x_1, y_1), u_2 = (x_2, y_2)$$, the line segment joining the points can be parameterized by the function $$l(t) = (tx_1 + (1-t) x_2, ty_1 + (1-t) y_2)= t u_1 + (1-t) u_2.$$ For example, if $$t = .25$$ it means we take $$25\%$$ $$u_1$$ and $$75\%$$ $$u_2$$. If the coefficients don't add to one you may leave the line segment.

Now, let $$u_1 = (x_1, f(x_1)$$ and $$u_2 = (x_2, f(x_2))$$. Look at this image taken from Wikipedia.

The requirement is that the curve of $$f(x)$$ lies below this secant line joining these two points, which is parametrized by $$l(t) = (tx_1 + (1-t) x_2, tf(x_1) + (1-t) f(x_2)),$$ the point indicated on the image for some particular $$t$$. The point $$(tx_1 + (1-t) x_2, f(tx_1 + (1-t) x_2)$$ on the curve needs to lie below it.

Another interpretation: $$tf(x_1) + (1-t) f(x_2))$$ is a weighted average of the outputs of $$f$$, while $$f(tx_1 + (1-t) x_2)$$ is output from taking a weighted average of the inputs. So you can say that the requirement for convexity is that

$$f(\text{weighted average of points}) \leq \text{weighted average of }f(\text{points}).$$

This generalizes to Jensen's inequality.

• "If the coefficients don't add up to $1$, you may leave the line segment." -> This one sentence actually answered a lot of questions I've been having with similar issues. Thank you!
– Sean
Mar 8, 2019 at 5:31
• No problem, glad to clear up confusion. Mar 8, 2019 at 5:37