$\textbf{Step 1:}$ $X_1 \cong Y_1 \quad \text{and} \quad X_2 \cong Y_2 \Rightarrow X_1 \oplus X_2 \cong Y_1 \oplus Y_2$ 1) Bijective: $g_{1}, g_{2}$ are bijective. 2) Isometry: Using that $g_{1}, g_{2}$ are Isometries it follows that $\Vert g(x,y) \Vert_{\ell^p} = \Vert (g_{1}(x), g_{2}(y)) \Vert_{\ell^p} = \left( \Vert g_{1}(x) \Vert_{X_{2}}^{p} + \Vert g_{2}(y) \Vert_{Y_{2}}^{p} \right)^{1/p} \\ = \left( \Vert x \Vert_{X_{1}}^{p} + \Vert y \Vert_{Y_{1}}^{p} \right)^{1/p} = \Vert (x,y) \Vert_{\ell^p}$

$\textbf{Step 2:}$ is shown here math.stackexchange.com/q/4909653/1318343 where I assumed that I is countable.

$\textbf{Step 3:}$ $\phi: \ell^p(I,X) \rightarrow \ell^p(I_1,X) \oplus \ell^p(I_2,X) \quad \text{defined by} \quad \phi(f) := (f|_{I_1}, f|_{I_2})$ is a isometric isomorphism

1. Bijective: Injective since $I_{1}$ and $I_{2}$ are disjoint and surjective since $I_{1} \cup I_{2}= I$ 2. Isometry: $\Vert \phi(f) \Vert_{\ell^p} = \Vert (f|_{I_1}, f|_{I_2}) \Vert_{\ell^p} = \left( \Vert f|_{I_1} \Vert_X^{p} + \Vert f|_{I_2} \Vert_X^{p} \right)^{1/p} \\ = \left( \left( \sum_{i \in I_1} \vert f(i) \vert^{p} \right) + \left( \sum_{i \in I_2} \vert f(i) \vert^{p} \right) \right)^{1/p} \\ = \left( \sum_{i \in I} \vert f(i) \vert^{p} \right)^{1/p} = \Vert f \Vert$

Thanks for your detailed answer! I genuinely appreciate it. I tried to prove the left gaps. Could you be so kind to review it? @DeanMiller

No problem! What you did for step 1 is fine if $X_{1}$ is isometrically isomorphic to $Y_{1}$ and if $X_{2}$ is isometrically isomorphic to $Y_{2}$, but that is not assumed in this case. Instead, you want to show that $g$ and $g^{-1}$ are both continuous using that $g_{1}$ and $g_{2}$ are linear homeomorphisms. For step 3, the isometry portion of the argument looks good, but $\phi$ is injective because $I_{1}$ and $I_{2}$ cover $I$ and is surjective because $I_{1}$ and $I_{2}$ are disjoint, so you had those conditions the wrong way around. Everything else looks good.

It is also worth briefly mentioning that you used $\| f\vert_{I_{1}} \|^{p}_{X}$ and $\| f\vert_{I_{2}} \|^{p}_{X}$ rather than $\| f\vert_{I_{1}} \|^{p}_{\ell^{p}(I,X)}$ and $\| f\vert_{I_{2}} \|^{p}_{\ell^{p}(I,X)}$ while showing that $\phi$ is an isometry in step 3. But I assume that is just a typo.

yes you are right. For the missing proof why $g$ and $g^{-1}$ are continuous: $g$ continuous as component wise continuity of $g_{1}$ and $g_{2}$. For the inverse I would just argue with the Open Mapping Theorem since $g$ is a linear bounded operator and bijective. is that right? @DeanMiller

That is right as to why $g$ is continuous. Applying the open mapping theorem or bounded inverse theorem to conclude that $g^{-1}$ is continuous is valid if all of the normed spaces in question are Banach spaces, but you can also do it without using those results by noting that $g^{-1}(x,y) = (g_{1}^{-1}(x), g_{2}^{-1}(y))$ for all $x\in Y_{1}$ and $y\in Y_{2}$ combined with an analogous argument as to the continuity of $g$. This way, the result is also valid in the normed space case as well and the proof uses results which are more elementary than the open mapping theorem.

that makes sense! thank you

you mentioned above that the internal and external direct sum are isomorph. I tried to prove this with $f(x,y) = x+y$ as the isomorphism. Linearity is clear. Surjective and Injective follows from the unique representation of every element in the internal direct sum. Nevertheless I am not sure about the continuity of both f and the inverse. Which norms do we use to prove this. Is it the mentioned "topological direct sum norm"? I hope my question is clear otherwise just ask! @DeanMiller

No problem. The inner direct sum and outer direct sum are algebraic concepts. What I mentioned in the first part of that post was simply discussing those concepts algebraic concepts. Then I mentioned the concept of a topological direct sum where you take two normed spaces and form an outer direct sum equipped with any one of those $\ell_{p}$ norms (which are all equivalent) to obtain another normed space.

I am not sure how it works here but I guess thats the chat?

ok so is my Ansatz right to show that they are isomorph?

What you did is correct to show that they are isomorphic in the vector space sense. There is also the notion of being isomorphic in the normed space or Banach space sense, where the linear bijection must also be a topological homeomorphism.

so what is missing to show that there isomorphic in the banach space sense? isnt it enough to show bijection and the continuity?

You can ask if given a Banach space $X$ and two closed subspaces $X_{1}$ and $X_{2}$ such that $X_{1} \cap X_{2} = \{0\}$ whether the topological direct sum $X_{1} \oplus X_{2}$ is isomorphic to $X_{1} + X_{2}$ as a vector subspace inheriting the normed space structure from $X$. I do not think this is true in general as there are examples of a Banach space $X$ and closed subspaces $X_{1}$ and $X_{2}$ such that $X_{1} + X_{2}$ is not a closed vector subspace of $X$.

I will need to think a bit more about whether there are nice conditions which imply that there is such an isomorphism.

If $X_{1}+X_{2}$ is closed in $X$, then you have that the map $f:X_{1}\oplus X_{2} \to X_{1}+X_{2}$ defined by $f(x,y) := x+y$ is a continuous bijective linear operator from one Banach space onto another Banach space. By the bounded inverse theorem, you can then conclude that $f$ is a Banach space isomorphism. Note that $X_{1}\oplus X_{2}$ equipped with any of the norms previously discussed is a Banach space whenever $X_{1}$ and $X_{2}$ are Banach spaces.

This also provides some information on why $X_{1} + X_{2}$ is not always closed: math.stackexchange.com/q/42993/990894

ok thank you, I can follow you. One thing I am still missing is the argument why it is continuous. Is it because with the previously discussed norms x+y is even an isometric isomorphism?

I don't think there is an isometric isomorphism here. To show continuity, you can use any of the previously discussed norms on $X_{1}\oplus X_{2}$ as they are all equivalent. One example is the norm given by $\|(x,y)\| := \max \{ \|x\|, \|y\| \}$. Then you have that $\| f(x,y) \| = \| x + y \| \leq \|x\| + \|y\| \leq 2 \|(x,y)\|$ for all $x\in X_{1}$ and $y\in X_{2}$. This shows that $f$ is continuous.

ok I will think about it, thank you!

No problem!

ok let me sum it op because I still have a problem with the boudness. Lets assume the following:

$ X = X_{1} \oplus_{i} X_{2}$ where I use the $i$ to denote the internal direct sum, the isomorphism is f((x,y)) = x+y as mentioned above. I want to use the above discussed $\ell^{p}$ norm for $p \in [1, \infty)$. So we have $\Vert f(x,y) \Vert_{X} = \Vert x +y \Vert_{X} \leq \Vert x \Vert_{X} + \Vert y \Vert_{X}$ but we actually need $\leq (\Vert x \Vert_{X}^{p} + \Vert y \Vert_{X}$)^{p})^{1/p}$ right?

You have that there exists some $C > 0$ such that $|a| + |b| \leq (|a|^{p} + |b|^{p})^{1/p}$ for all $a,b\in\mathbb{R}$ because the $\ell_{1}$ and $\ell_{p}$ norms are equivalent on the finite-dimensional normed space $\mathbb{R}^{2}$. Then your desired inequality follows by applying this with $a = \| x \|_{X}$ and $b = \| y \|_{X}$. In fact, you can take $C$ as $2^{1 - 1/p}$ for a best possible estimate.