Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question
add comment

1 Answer

up vote 2 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.

share|improve this answer
    
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
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.