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The following construction appears to yield a dense Graph in $H\times H$ where $H$ is a seperable Hilbert-space.

Take $\{x_n\}$ a countable dense subset of $H$. Let $\{e_n\}$ an orthonormal basis of $H$, and define $Te_n :=x_n$. Then extend $T$ linearly to $D(T)$.

Then $G(T)$ is dense. Why?

How can we construct a sequence $(x_k, Tx_k)$ such that we reach every $(x,y) \in H\times H$. Or is there a more straightforward approach? Any ideas, or suggestions are welcome. (It is Rudin Excercise 13.3)

DinkyDoe
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2 Answers2

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Let $(x,y)\in H\times H$ and $\varepsilon>0$. Clearly, there exist $a_1,\ldots,a_n$, s.t. $$ \|a_1e_1+\cdots+a_ne_n-x\|<\varepsilon. $$ Let $z=y-(a_1x_1+\cdots+ a_nx_n)$. As $\{x_n\}$ is dense in $H$, there is an $m>n$, such that $\|x_m-z/\varepsilon\|<1$. Let $w=z-\varepsilon x_m$, then $\|w\|<\varepsilon$ and $$ T(a_1e_1+\cdots+a_ne_n+\varepsilon e_m)=T(a_1e_1+\cdots+a_ne_n)+\varepsilon Te_m=y-z+\varepsilon x_m=y-w, $$ and hence $$ \|T(a_1e_1+\cdots+a_ne_n+\varepsilon e_m)-y\|=\|w\|<\varepsilon. $$ Altogether $$ \big\|\big(x,y\big)-\big(a_1e_1+\cdots+a_ne_n+\varepsilon e_m, T(a_1e_1+\cdots+a_ne_n+\varepsilon e_m)\big)\big\| < 3\varepsilon. $$ QED

Yiorgos S. Smyrlis
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  • that looks neat. But I dont quite grasp why you could take a different $\epsilon´$ and everything still works..why is this $\epsilon´$ such that $||a_1e_1+\cdots +a_ne_n+\epsilon´e_m - x|| < \epsilon$? Thatś what you use right? Iḿ little puzzled by this change of epsilon. – DinkyDoe Dec 26 '13 at 23:55
  • I guess we want to add up a little bit of (smaller) $\epsilon x_m$ to $a_1e_1 + \cdots + a_ne_n$ so that the difference in norm with $x$ is still sufficiently small. I guess that is possible, but I am not 100% sure why. Maybe its obvious. – DinkyDoe Dec 26 '13 at 23:59
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    Actually, you are right. No need for an additional $\varepsilon$ - Correction done. – Yiorgos S. Smyrlis Dec 27 '13 at 00:01
  • Ok, thatś very neat. I see now. Thank you kindly for the effort. – DinkyDoe Dec 27 '13 at 00:03
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First of all, $T$ is definable only in a dense subset of $H$, as $$ T\Big(\sum_{n=1}^\infty a_ne_n\Big)=\sum_{n=1}^\infty a_nx_n, $$ is definable only if the sum in the right hand side converges absolutely in the $H$-norm. But it is definable for finite sums, which are dense.

Yiorgos S. Smyrlis
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  • Sorry, I know that $D(T)$ is dense in $H$, which like you mentioned can only contain finite sums. The actual idea of showing that $G(T)$ is dense is not clear to me at all. The $H\times H$-norm is $||(x,y)||² = ||x||²+ ||y||² $. It is not obvious at all why we can find a converging sequence in $H\times H$-norm so that $(x_n, Tx_n)\to (x,y) \in H \times H$ – DinkyDoe Dec 26 '13 at 21:57