Prove that there exist an Hilbert space $H$ and a linear continuous operator $F: H \mapsto H$ such that: $$(*) \ \ \ \ (Fu,u) > 0 \qquad \forall u \in H \setminus \{ 0 \}, $$ but $F$ is not invertible. Suppose $H=\mathbb{R}^n$, why is it now impossible to find such $F$? A hint is given: consider $H=l^2$ and $F$ defined as $Fx=a_n x_n$ if $x \in l^2$, for a suitable sequence $(a_n)_n$.
I think the answer to the last question is that $\dim \mathbb{R}^n < \infty$. Back to the first problem, suppose we found a (positive) sequence $a_n$ such that $F$ satisfies $(*)$ for every $x \in l^2$, but then how can $F$ be non-invertible? If $F$ is not invertible then there exists $x \in l^2$ such that $x \neq 0$ and $Fx=0$, which contradicts $(*)$. Something is wrong right here, it seems like I'm running in circle..
$F:\ell^2(\mathbb{N})\rightarrow \ell^2(\mathbb{N})$ defined by $(Fx)_j=\frac{1}{j}x_j$ is well defined linear and bounded and fulfills $(\ast)$ but is not surjective since for example $\{\frac{1}{j}\}$ is not in the range of $F$.