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I want to prove that the direct sum of two (complex) Hilbert spaces is a Hilbert space. I've shown that we have an inner product and also shown norm however I have trouble to show converges. We define our Hilbert spaces as follows, let $H_1,H_2$ be Hilbert spaces, then the direct sum $H_1\oplus H_2$ is defined below $$\langle (x,y)|(x',y')\rangle:=\langle x|x’ \rangle_{H_1}+\langle y|y’ \rangle_{H_2} ,$$ where $(x,y),(x’,y’)\in H_1\times H_2 $.

Yet I have readed a proof for a similar question however I didn't get so much from it. Here is the link. Countable family of Hilbert spaces is complete

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  • $\begingroup$ Do you mean $(h_1,k_1) = (x,y)$ ? It seems you used $2$ diferent notations. $\endgroup$
    – RicardoMM
    Nov 17, 2022 at 18:51
  • $\begingroup$ Sorry yes! I will edit that. $\endgroup$
    – NabbKitha
    Nov 17, 2022 at 18:52
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    $\begingroup$ If you write the definition of Cauchy Sequence aplied to your elements it shouldn't be too hard to figure out - you are adding 2 numbers that converge. $\endgroup$
    – RicardoMM
    Nov 17, 2022 at 19:02

1 Answer 1

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Say you have a sequence $(z_n) $ that is Cauchy in $H=H_1 \oplus H_2$. Then $ \exists! (x_n) \in H_1$ and $ \exists !(y_n) \in H_2$ such that $ \forall n$ we have $z_n=x_n +y_n$. Now note that if $z= x+y $ where $x \in H_1$ and $ y \in H_2$ then $ (x, y) = 0$ (Check this by writing $x= x+ 0_{H_2}$ and $y=0_{H_1} +y $) $$ \| z_n \| ^2 = \| x_n \|^2+ \| y_n \|^2 $$

This will be enough to show that both $ (x_n) $ and $(y_n)$ are Cauchy. These being Cauchy, implies they are convergent. say $$ (x_n) \to \bar{x} $$ $$ (y_n) \to \bar{y} $$ then one can show convergence of $z_n$ i.e. $$z_n \to \bar{x} + \bar{y}$$ using triangle inequality ( $ ( \| z_n \| \leq \| x_n \| + \| y_n \|$ )

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