Is it covex function?$J_{new}(u)=\int_{\Omega} \sum_{i=1}^{N} \lambda_if(x)u_i(x)dx$

Question

I have a function such as $$J(u)=\int_{\Omega} \sum_{i=1}^{N} f(x)u_i(x)dx$$ where $f(x):\Omega \to R$, $0 \le u_i(x) \le 1,\sum_i u_i(x)=1$

Given that $J(u)$ is a convex function w.r.t $u$. Now I will multiple my $J(u)$ function with coefficient $\lambda=[\lambda_1..\lambda_N]$ such as $$J_{new}(u)=\int_{\Omega} \sum_{i=1}^{N} \lambda_if(x)u_i(x)dx$$

where $0 \le \lambda_i \le 1$.

My question is that $J_{new}(u)$ is whether convex or non-convex function w.r.t $u$? If it is convex function w.r.t $u$, please prove it? Thanks in advance

I think that it is still convex function w.r.t $u$

Answer 1

The argument $u$ is linear in $J$ and $J_{new}$.