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This is a question about (possible unbounded) operators. We know that $\mathcal{D}(T^*)=\{0\}$ iff $\mathcal{G}(T)$ is dense in $\mathcal{H}\times\mathcal{H}$, where $\mathcal{H}$ is a separable Hilbert space and $\mathcal{G}(T)$ the graph of an operator $T$ with domain $\mathcal{D}(T)$. My task is to show that this equivalence really can happen by contsruction of an operator $T$.

I have done this so far: $\mathcal{H}$ is separable thus there exists an orthonormal countable base $(e_n)_{n\in\mathbb{N}}$. We can also find $(x_n)_n$ an countable dense subset of $\mathcal{H}$ by assumption. Define $T(e_n)=x_n$ and extend this by linearity to a domain $\mathcal{D}(T)$ which is defined to be the set of all finite linear combinations of $(e_n)_n$. I can show the following: $\mathcal{D}(T)$ is dense in $\mathcal{H}$. Thus we can apply the theory of unbounded operators to conclude the existence of the adjoint $T^*$ on $\mathcal{D}(T^*)$. But can we know conclude from out this directly that $\mathcal{G}(T)$ is dense in $\mathcal{H}$, that $\mathcal{D}(T^*)=\{0\}$ and thet $T^*$ is the zero-operator (the last to facts are my main problem)? Can we also conclude the other direction from out this?

I would be happy about help and solutions $:)$

Thank you

domainTst
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  • I'm discussion this exercise here http://math.stackexchange.com/questions/621482/rudin-13-3-zero-operator-as-adjoint – simon Dec 30 '13 at 13:23

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Hint: for any $(x,y)$ in ${\mathcal H} \times {\mathcal H}$ and $\epsilon > 0$, first take $u \in {\mathcal D}(T)$ that is close to $x$, then add a small multiple of a suitable $e_n$ to get $v$ where $T v$ is close to $y$.

Robert Israel
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  • This can be by definition of $T$ right? Let $y\in\mathcal{H}$ then there exists $x_n$ in an $\varepsilon$-nbhd of $y$ since $(x_n)_n$ is the set which is dense in $\mathcal{H}$, thus take $Te_n$ this is b definition $x_n$ and thus also in such a nbhd. But why is $\mathcal{D}(T^)=0$ and $T^=0$, thats my main problem, mayby i have to state this. – domainTst Dec 20 '13 at 18:07