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I was doing a proof and I need to show a result to conclude it:

$X$ is a reflexive Banach space with a norm, $\|\cdot\|$, of class $\mathcal{C}^1$.

$f:X\to\overline{\mathbb{R}}$ is lower semicontinuous and convex.

If $\lambda>0$ and $x\in X$ are prefixed, I define $\Phi:X\to\overline{\mathbb{R}}$ such that $\Phi(y)=f(y)+\frac{1}{2\lambda}\|x-y\|^2$.

Then I need to show that $\Phi$ is coercive, this is: $\displaystyle\lim_{\|y\|\to\infty} \Phi(y)=+\infty$.

Maybe the result isn't true... Thanks.

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Perhaps you need to explain what is C1? – cheng Jan 23 '12 at 10:15

Hint 1: The triangle inequality implies that $\|y\| - \|x\| \leq \|y-x\|$.

Hint 2: Convexity implies that there exists a linear functional $\ell:X\to\mathbb{R}$ and a real number $r$ such that $f(y) \geq \ell(y) + r$.

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Yeah, but the linear function $\ell$ implies that the square of the norm is stronger (in the sense of that implies the coercivity of $\Phi$)? – Nerea Jan 24 '12 at 23:33
Yes, by definition a continuous linear functional satisfies $-C \|x\| \leq \ell(x) \leq C \|x\|$ for some fixed constant $C$. The existence of $\ell$ follows from (though not using the full force of) this paper, where it is shown that for a lcs convex function $f$ on a Banach space $X$, the set on which $f$ is subdifferentiable is dense in $\{ f < \infty\}$. (If $\{f <\infty\} = \emptyset$, your claim is trivial.) – Willie Wong Jan 25 '12 at 8:40
Ahmm, many thanks – Nerea Jan 25 '12 at 17:02

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