Given $f: X \to \mathbb{R}$ a strict convex, lower semi-con. function with $X$ reflexive.
Show that the inclusion
$$l \in\partial f(x) \mbox{ (subgradient of f)}$$
posses an unique solution for every $l\in X^*$, under the assumption $\frac{f(x)}{\|x\|} \to \infty$.
My Idea is to show that the minimizer of $f(x)-l(x)$ is the unique solution. But is that possible?
Let $l \in X^*$, then we see that $f-l$ is strictly convex, lsc. and $\lim_{\|x\| \to \infty} {f(x)-l(x) \over \|x\|}= \infty$.
If we let $\phi= f -l$ we see that since $\partial \phi(x) = \partial f(x) - \{l\}$, then $0 \in \partial \phi(x)$ iff $l \in \partial f(x)$. Hence we can just deal with showing that $0 \in \partial f(x)$ for a unique $x$.
Consider $L_\alpha = \{x | f(x) \le \alpha \}$. Let $\alpha_n = { 1\over n}+\inf f$, then $L_{\alpha_n}$ are nested, non empty weakly compact sets (since $X$ is reflexive). Hence $L=\cap_n L_{\alpha_n}$ is non empty and hence $f$ has a minimiser, so we have $0 \in \partial f(x)$ for some $x \in L$.
Now suppose $x,y $ are such that $0\in \partial f(x)$, $0\in \partial f(y)$, then $tx+(1-t)y$ is a minimiser for $t \in [0,1]$ which contradicts strictness.