Consider the problem (P)
$\min f(x)$
s.t $g(x)\leq0$
$x\in X$
Where $f,g$ is convex and $X$ is convex. Suppose that $x^*$ is optimal solution of the problem above and $g(x^*)<0$ Show that $x^*$ is optimal solution of the problem below (P')
$\min f(x)$
s.t $x\in X$
I've thought that $x^*$ is in the set of optimal solution of P and we can see that $P\subseteq P'$ and therefore $x^*\in P'$ and is optimal solution of $P'$ but I am not sure if this is the right way.
We've just started convex optimization
Any if anyone can give some geometric intuition it will be nice
Let us assume the $x'$ minimize $(P')$.Looking at the interval $[x',x^*]\subseteq X$ because $X$ is convex. one can find $\lambda\in[0,1]$ such that $g(\lambda x'+(1-\lambda)x^*)\leq\lambda g(x')+(1-\lambda)g(x^*)\leq g(x^*)< 0$.\ Now looking at $f(\lambda x'+(1-\lambda)x^*)\leq\lambda f(x')+(1-\lambda)f(x^*)\leq f(x^*)$ and we got new minimizer for $(P)$ in contradiction that $x^*$ is the minimizer