I have a function $f(x)$ defined as follows $$f(x)=f_1(x_0,x_1)+f_2(x_0,x_2)+\cdots f_T(x_0,x_T)$$ each of the individual function $f_i(x_0,x_i)$ is non-negative and jointly concave over $(x_0,x_i)$ therefore the summation is also concave. If I fix $x_0=c$ then the new function becomes $$f_c(x)=f_1(c,x_1)+f_2(c,x_2)+\cdots f_T(c,x_T)$$ My question is as follows. Is $f_c(x)$ a concave function of $x_1,x_2 \cdots x_T$? Any help in this regard will be much appreciated. Thanks in advance.


\begin{align} f_c(\lambda x + (1-\lambda) y)&=\sum_{i=1}^Tf_i(\lambda c+(1-\lambda)c, \lambda x_i + (1-\lambda) y_i)\\ &\geq \sum_{i=1}^T [\lambda f_i(c,x_i)+(1-\lambda)f_i(c,y_i)] \\&=\lambda f_c(c,x) + (1-\lambda)f_c(c,y) \end{align}

  • $\begingroup$ So means $f_c(x)$ will be individually concave function?\ $\endgroup$ – Frank Moses Feb 5 '18 at 1:03
  • $\begingroup$ what does "individually concave function" mean? Do you mean if all other variables are fixed, is it still a concave function? answer is yes if that is what you mean. $\endgroup$ – Siong Thye Goh Feb 5 '18 at 1:10
  • $\begingroup$ I am very sorry for asking this. But can you please provide your help in understanding this problem. (I am sorry again for directing your attention to another question but I will be very thankful for your help) math.stackexchange.com/questions/2636419/… $\endgroup$ – Frank Moses Feb 5 '18 at 1:32

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