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Let $(X_i, ||\cdot||_{X_i})$,  $i=1, 2$, be Banach spaces. Consider the direct product space $X:=X_1\times X_2$ with usual component-wise operations. We know we can endow this space with a norm (one possible choice is $||\cdot||_1:=||\cdot||_{X_1}+||\cdot||_{X_2}$).  
Let us suppose we have chosen some norm and that we keep it fixed. Denote this norm with $||\cdot||_X$.  Are the following claims true?

Claim 1.  For any $(x_1, x_2)\in X$ one has $||(x_1, x_2)||_X\geq ||x_i||_{X_i}$,  $i=1, 2$.

I know this claim is true for "usual" norms on $X$ like $||\cdot||_1, ||\cdot||_2$ and $||\cdot||_{\infty}$ but I don't seem to be able to prove it for an arbitrary norm.

Claim 2.  Let $\{(x_n,y_n)\}_{n=1}^{\infty}$ be some sequence in $X$. Then $(x, y)=\lim_{n\to\infty}(x_n, y_n)$ if and only if $x=\lim_{n\to\infty} x_n$ in $X_1$ and $y=\lim_{n\to\infty} y_n$ in $X_2$.

I've found one part of the "answer" for this last claim here:   Does convergence of vector sequence imply that of all components?. However, I am not exactly satisfied with it as I fail to see why the projections should have to be continuous in the topology of this, arbitrary, norm on $X$.
Also, I am aware that one direction in the second claim follows immediately from Claim 1 (should it be true).

I appreciate any help I can get!

forbes
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1 Answers1

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For the first one, let $X_{1}=X_{2}=:Y$, we can create $\|(x_{1},x_{2})\|_{X}=\dfrac{1}{3}\|x_{1}\|_{Y}+\dfrac{1}{3}\|x_{2}\|_{Y}$, this is a norm in $X$., while $\|(x,x)\|_{X}=(2/3)\|x\|_{Y}<\|x\|_{Y}$ for $x\ne 0$.

For the second one. If $X$ is endowed with the usual product topology, choose a subbasic open set in $X$, it is of the form $G\times H$ for open $G$ in $X_{1}$, $H$ open in $X_{2}$, by this observation then it is true.

If $X$ is endowed with other topology, then the answer is negative. Consider $X_{1}=X_{2}=L^{1}(\mathbb{R})\cap L^{2}(\mathbb{R})$. We let $\|(f,g)\|_{X}=\|f\|_{L^{1}(\mathbb{R})}+\|g\|_{L^{1}({\mathbb{R}})}$, where we endow $X_{1}$ and $X_{2}$ both the $L^{2}(\mathbb{R})$ norm.

user284331
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  • I don't understand why your norm is actually a norm. You have $$ \Vert (0, x_2)\Vert_X = 0$$ but we could have $x_2\neq 0$. – Severin Schraven Oct 27 '19 at 18:42
  • I guess you wanted to write $$\Vert (x_1, x_2)\Vert_X =\frac{1}{2} (\Vert x_1\Vert_{X_1} + \Vert x_2\Vert_{X_2})$$ – Severin Schraven Oct 27 '19 at 18:53
  • What's wrong to have instead the one-third? – user284331 Oct 27 '19 at 18:55
  • The homogeneity and triangle inequality still hold, no? – user284331 Oct 27 '19 at 18:58
  • Sorry, did not reload the page and was wondering what you meant – Severin Schraven Oct 27 '19 at 18:59
  • Of course a third is fine. – Severin Schraven Oct 27 '19 at 19:00
  • I'm sorry, I don't quite understand the second part of your answer: I agree that projections are continuous in standard product topology. However, maybe I was unclear, but I meant that we start with some norm on $X$ (we don't know which) and we are considering topology induced by it. Why should the projections be continuous in this topology? – forbes Oct 27 '19 at 19:13
  • That's why I come out with a counterexample. – user284331 Oct 27 '19 at 19:14
  • Okay, forgot to reload the page. One last thing, can you give me an example of a sequence $(f_n)\subseteq L^{1}(\mathbb{R})\cap L^{2}(\mathbb{R})$ such that $f_n\to f\in L^1\cap L^2 $ in $L^1$ but $f_n\nrightarrow f$ in $L^2$. Anyhow, thanks a lot for your insights! – forbes Oct 27 '19 at 19:42
  • Take $f_{n}=n^{-1}\chi_{[0,n]}$, then $f_{n}\rightarrow 0$ in $L^{2}$ but not in $L^{1}$. – user284331 Oct 27 '19 at 19:50
  • Take $f_{n}=n^{1/2}\chi_{[0,1/n]}$, then $f_{n}\rightarrow 0$ in $L^{1}$ but not in $L^{2}$. – user284331 Oct 27 '19 at 19:53