Show that $J\colon V \to \mathbb{R}$ is convex

Question

Let $V$ be a real, reflexive Banach space and $a\colon V \times V \to \mathbb{R}$ a symmetric, positive definite bilinear form and the energy function $J\colon V \to\mathbb R$ given by $$J(v):={1 \over 2}a(v,v)-<f,v>$$

How can I show that $J$ is convex, that means that for all $u,v \in V$ where $u \neq v$ and $\lambda \in (0,1)$ we have $J(\lambda u +(1-\lambda)v) < \lambda J(u) + (1-\lambda) J(v)$? A hint would be much appreciated.

Answer 1

A convex function added to a linear function is convex. Hence it is sufficient to show that $x \mapsto a(x,x)$ is convex.

It suffices to show that this is true on any line $t \mapsto x_0 + t d$ Since $p(t)=a(x_0+td, x_0+td) = a(x_0,x_0) + 2t a(x_0,d) + t^2a(d,d)$, we see that $p$ is convex and, since $x_0,d$ were arbitrary, so is $a$.

Answer 2

Let $u,v\in V$ be fixed with $u \neq v$. We have:

\begin{align} J\Big(\lambda u + (1-\lambda)v\Big) = &\,\,\,\frac{\lambda^2}2 \,a(u,u) + \lambda(1-\lambda) \,a(u,v) + \frac{(1-\lambda)^2}2 \,a(v,v)\\ -&\,\,\,\Big( \lambda \langle f, u\rangle + (1-\lambda)\langle f,v\rangle \Big). \end{align}

On the other hand, we have $\lambda\, J(u) = \frac{\lambda}2\, a(u,u) - \lambda\langle f, u\rangle$ and $(1-\lambda)\, J(v) = \frac{(1-\lambda)}2\, a(v,v)\,-$ $\lambda\langle f, v\rangle$. It follows that $J\Big(\lambda u + (1-\lambda)v\Big) - \lambda\, J(u) - (1-\lambda)\, J(v)$ equals

$$g(\lambda) = \frac{\lambda^2-\lambda}2\,a(u,u) + \lambda(1-\lambda) \,a(u,v) + \frac{(1-\lambda)^2-(1-\lambda)}2 \,a(v,v),$$

where $\lambda \in [0,1]$. Notice that $g(0)=g(1)=0$ and that $g$ is a polynomial of degree $2$ in $\lambda$ whose leading coefficient is

\begin{align} \frac12\,a(u,u) - a(u,v) + \frac12\,a(v,v) &= \frac12\,a(u-v,u-v) .\end{align}

Because $a$ is positive definite and $u\neq v$ we have $u-v\neq 0$ and hence the leading coefficient is positive. It follows that $g<0$ between its roots, that is, $g(\lambda)<0$ for $\lambda \in (0,1)$.