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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?

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up vote 3 down vote accepted

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.

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Sorry, but I can't understand why $2\langle a,u\rangle+b+\|u-v\|^2=\|u-v+a\|^2+b+2\langle a,v\rangle-\|a\|^2$. THe space is not Hilbert. – Tomás Feb 28 '13 at 20:20
Expand $\|u-v+a\|^2=\langle u-v+a,u-v+a\rangle=\|u-v\|^2+2\langle a,u-v\rangle+\|a\|^2$. – Did Feb 28 '13 at 20:24
$X$ is a non-reflexive Banach space, it is not Hilbert. The symbol $\langle\cdot,\cdot\rangle$ represents duality in general. – Tomás Feb 28 '13 at 21:38
Thank You @did. – Tomás Mar 1 '13 at 12:23
You are welcome. Thanks for nudging me. – Did Mar 1 '13 at 16:18

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