Let $a \in \mathbb R^n$ be a fixed vector. Define $f(x)$ on $\mathbb R^n$ by $f(x)=a\cdot x$
1) Show that $f(x)$ is not coercive.
1) Show that if $\epsilon >0$ , then $g(x)=f(x)+\epsilon ||x||^2$ is coercive.
How can I prove this ? So far I'm stuck without a single clue.
If you take the limit of $f$ on the direction that is orthogonal to $a$, say $x=\gamma a^{\perp}$ for $\gamma\to\infty$, then $$ \lim_{||x|| \to \infty} f(x) =0. $$ This means that the limit is not always infinity and varies based on the direction (actually it does not exist!).
Consider the function $g$ and see that: $$ g(x)=f(x)+\epsilon \|x\|^2=\|\sqrt\epsilon x+\frac{a}{2\sqrt{\epsilon}}\|^2-\frac{\|a\|^2}{4\epsilon}. $$ See that : $$ \|\sqrt\epsilon x+\frac{a}{2\sqrt{\epsilon}}\|^2\geq \epsilon \|x\|^2-\frac{\|a\|^2}{4\epsilon}. $$ And taking the limit shows what you want.