# Direct sum of two Hilbert spaces is a Hilbert space

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

• Do you mean $(h_1,k_1) = (x,y)$ ? It seems you used $2$ diferent notations. Nov 17, 2022 at 18:51
• Sorry yes! I will edit that. Nov 17, 2022 at 18:52
• 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. Nov 17, 2022 at 19:02

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 ( $$( \| z_n \| \leq \| x_n \| + \| y_n \|$$ )