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Suppose I have a multiple variable function $f(X,Y) = \sum_i \sum_j k_{ij} g(x_i,y_j) + \sum_j h(y_j)$ where $X = \{x_i\}$ and $Y = \{y_j\}$. $k_{ij}$ is a non-negative constant. I want to show whether $f(X,Y)$ is convex. It is difficult to directly show the convexity of $f(X,Y)$, so I attempt to prove that the components $g(x_i,y_j)$ and $h(y_j)$ are both convex in their domains. Does it imply $f(X_,Y)$ is also convex since $f(X,Y)$ is a linear combination of $g(x_i,y_j)$ and $h(y_j)$?

JYY
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  • As long as the coefficients are nonnegative, yes. Note that splitting up $X$ and $Y$ isn't important. – Ben W Feb 11 '20 at 03:34
  • I "feel" that $f(X,Y)$ is convex. The way I check the convexity of $g(x_i,y_j)$ is to evaluate the signs the pivots of its Hessian. Since $g(x_i,y_j)$ is a two-variable function, its pivots are easily computed. $f(X,Y)$ has $|X|+|Y|$ variables, its Hessian is hard to work with. – JYY Feb 11 '20 at 03:38
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    Unless I'm missing something, you're making this waaaay too complicated. A convex function is one where $f(t\textbf{x}+(1-t)\textbf{y})\leq tf(\textbf{x})+(1-t)f(\textbf{y})$. Suppose $f(\textbf{x})$ is a nonnegative linear combination $\sum k_ig_i(\textbf{x})$ of convex functions $g_i$. Then $f(t\textbf{x}+(1-t)\textbf{y})=\sum k_ig_i(t\textbf{x}+(1-t)\textbf{y})\leq\sum k_i[tg_i(\textbf{x})+(1-t)g_i(\textbf{y})]=t\sum k_ig_i(\textbf{x})+$... well, you get the idea.. – Ben W Feb 11 '20 at 04:05
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    A non negative multiple of a convex function is convex. The sum of convex functions is convex. That's all. – copper.hat Feb 11 '20 at 04:15

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