Is it convex function?

Question

I have a function and I don't know it is whether convex or non-convex: $$J(c,\alpha)=\int_\Omega ( \alpha c-I(x))^2u \, dx+ \|\alpha\|^2$$ where $0 \le u \le 1$, $I(x): \Omega \to R$, $c$ is constant

Is it convex function? As my knowlegde, it is convex function because $\int_\Omega ( \alpha c-I(x))^2u \, dx$ is convex function in term of $0 \le u \le 1$ and $ \|\alpha\|^2$ is also convex, then totally, $J(c,\alpha)$ is convex.

Answer 1

As a function of $c$, it is convex. As a function of $\alpha$, it is convex. But it is not jointly convex in $\alpha$ and $c$.

Answer 2

If $x \mapsto f(x,t)$ is convex for each $t$, and $\mu$ is a positive measure, then $x \mapsto \int f(x,t) d\mu(t)$ is convex.

It follows that convex combinations of convex functions are convex.

Hence $\alpha \to J(c,\alpha)$ is convex.

In particular, since $(\alpha,x) = ( \alpha c-I(x))^2u$ is convex for each $x$, then $(\alpha,x) = \int_\Omega ( \alpha c-I(x))^2u dx $ is convex (using $\mu = \lambda$, the Lebesgue measure).

The function $\alpha \mapsto \|\alpha\|^2$ is convex.

The sum of convex functions is convex.