Let $X$ be a non-reflexive Banach space and $f:X\rightarrow\mathbb{R}$ a $C^1$ function that is Strongly Convex, i.e. $$f(u)-f(v)\geq\langle f'(v),u-v\rangle+c\|u-v\|^2$$
where $c>0$ is constant.
Is it possible for $f$ to be unbounded below?
[Math] Is it possible for a Strongly Convex function to be unbounded below
banach-spacesconvex-analysisfunctional-analysis
Best Answer
No. For every $a$, $b$ and $v$, the function $g:u\mapsto 2\langle a,u\rangle+b+\|u-v\|^2$ is bounded below. To wit, $$ g(u)=\|u-v+a\|^2+b+2\langle a,v\rangle-\|a\|^2\geqslant b+2\langle a,v\rangle-\|a\|^2. $$ Apply this to $a=f'(v)/(2c)$ and $b=(f(v)-\langle f'(v),v\rangle)/c$.
Edit: In a Banach space $X$, note that if $f'(v)$ is in the continuous dual space $X'$, there exists a finite $k$ such that $\langle f'(v),u-v\rangle\geqslant k\|u-v\|$ for every $u$ hence $f(u)\geqslant f(v)+\min\{kt+ct^2\mid t\geqslant0\}$. Since $c\gt0$, the RHS is finite and $f$ is uniformly bounded from below.