Is this a valid proof? If not what is missing or wrong? or an unnecessary assumption?

If $f(u)$ is a convex function, one has \[ f(\bar u)=f(\sum_{i=1}^Np_iu_i)\leq\overline{f(u)}=\sum_{i=1}^Np_if(u_i), \] where $\sum_{i=1}^Np_i=1$ and $p_i\geq0$, $\forall i$.

From elementary geometry, one knows that the point $(\bar u, \overline{f(u)})=\sum_{i=1}^N(u_i, f(u_i))p_i\in P$, where $P$ is the convex polygon whose vertices are the set of points on the plane $\{(u_i, f(u_i))\vert i=1, \dots, N\}$. Here, we assume that the sequence of $u_i$ is in a monotonically growing order, if $i<j$ then $u_i<u_j$. The point $(\bar u, \overline{f(u)})$ is the center of mass of the particles of mass $p_i$ located at the vertices of $P$. As $f(\cdot)$ is a convex function, by definition, $f(\lambda u_i+(1-\lambda)u_j)\leq \lambda f(u_i)+(1-\lambda )f(u_j)$, with $0\leq\lambda\leq 1$. Consequently, as $\inf \{U\}\leq\bar u\leq \sup\{U\}$, where $U=\{u_i\vert i=1, \dots, N\}$. Since one can always find an $u_i\in U$, such that $u_i\leq\bar u\leq u_{i+1}$, the point $(\bar u, f(\bar u))$ is outside and below the polygon $P$. Therefore $f(\bar u)\leq\overline{f(u)}$.



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