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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..

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    Invertible here means continuously invertible. For $F$ to be invertible you need $(a_n)\in l^\infty$. For $F^{-1}$ to be continuous you need that $|a_n|$ is bounded from below by a strictly positive bound. As you have said, for $\mathbb{R}^n$ this can't happen due to the finite-dimension but you should elaborate this a bit more ;) – 2017-01-14
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    Ok, thank you. So rather then focus on the "there exists $x$ such that $Fx=0$..., I should focus on showing that $F^{-1}$ cannot be continuous, right? For the finite dimensional question, maybe it suffices to say that linearity is equivalent to continuity. – 2017-01-14
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    With the given condition $F$ is always injective. So what goes wrong is surjectivity. If it where surjective, then by the CIT (as we have Hilbert spaces) the inverse would be continuous as-well. For this kind of multiplication operator we know what the inverse needs to be and in some sense the unboundedness of the inverse shows non-surjectivity: desired preimages aren't in $l^2$ anymore. For the finite-dimensional case, injective is equivalent to bijective is equivalent to continuously invertible (since everything there is continuous). – 2017-01-14

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$F:\ell^2(\mathbb{N})\rightarrow \ell^2(\mathbb{N})$ 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 $F$.