Is the logarithm of sum of multiple variables with the constraint on sum of them Concave?

Question

I know that without any constraint $log \sum_{i=1:1:m} \alpha_i C_i $ is not Concave but I am wondering is this function Concave when we have the constraint that $ \sum_{i=1:1:m} \alpha_i =1 $ and they are positive? C is simple constant vector (with all non-zero elements).

Answer 1

There is a bit of a terminology issue here. The status of a function as convex or concave is not affected by whether or not you "have constraints". You are talking about the context of convex optimization, though, so I think we can affirmatively answer the question I think you're trying to ask, and in the process correct the terminology.

As I said in the comments, your function $f(\alpha)=\log\sum_i C_i \alpha_i$ is concave. This means that the optimization problem $$\text{maximize} ~ f(\alpha)$$ is convex. Again: the function is concave, the optimization problem is convex. That's because convex optimization problems can either involve 1) minimizing a convex function or 2) maximizing a concave function. I do not get to say that my concave function $f$ is convex just because I'm using it in a valid convex optimization model. (If the machine learning community is doing this, they need to stop. Lead the charge!)

Now, of course, an optimization model can "have" constraints, as you say. In your case, you seem to want this: \begin{array}{ll} \text{maximize} & \log \sum_{i=1}^m C_i \alpha_i \\ \text{subject to} & \sum_{i=1}^m \alpha_i = 1 \\ & \alpha_i \geq 0, ~ i=1,2,\dots, m \end{array} This remains a convex optimization problem, because 1) I'm maximizing a concave objective and 2) all of my constraints are individually convex. But again, the function $f$ is still concave.

If you're really determined to stick to functions, then what you might say is that the domain of the function $f$ is artificially constrained. For instance: $$f(\alpha) \triangleq \sum_i \log_{i=1}^m C_i\alpha_i, \quad \mathop{\textrm{dom}} f \triangleq \left\{\alpha\in\mathbb{R}^m\,|\,\alpha\succeq 0, \vec{1}^T\alpha=1\right\}$$ In this case we've defined the domain of the function to be limited to the simplex using the constraints. A necessary condition for a function to be concave is that its domain is a convex set, which is satisfied here.

Answer 2

The answer is no (if I understand the question correctly). The function $$\log(x +1), $$ which is not convex, is of the form you gave with $$ \begin{array}{cccc} x=\alpha_1 & 1-x=\alpha_2, & C_1=2 , & C_2=1. \end{array} $$