Given $X,Y$ are Banach spaces with norms $\|x\|_X,\|y\|_Y$, prove $\|(x,y)\|=\max(\|x\|_X,\|y\|_Y)$ is a norm and defines a Banach space

Question

Here is the question:

Let $X$ and $Y$ be Banach spaces with norms $\|x\|_X$ and $\|y\|_Y$ respectively. Prove that

$$\|(x,y)\|=\max\{\|x\|_X,\|y\|_Y\}$$

defines a norm on $X\times Y$, and that $X\times Y$ with the norm is a Banach space.

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So this is what I have so far. I am hoping to get some input on it, especially the Banach space proof:

Norm proof

1) Given $x\in X$ and $y\in Y$, as $\|x\|_X$ and $\|y\|_Y$ are norms, we know that $\|x\|_X \geq 0$ and $\|y\|_Y\geq 0$, and $\|x\|_X=0$ if and only if $x=0$, and $\|y\|_Y=0$ if and only if $y=0$. Therefore it is obvious from definition that $\|(x,y)\|\geq 0$. If $x=0$ and $y=0$ then $\|(0,0)\|=\max\{\|0\|_X,\|0\|_Y\}=\max\{0,0\}=0$. If $x\neq 0$ and $y=0$, then $\|x\|_X\gt 0$ and $\|y\|_Y=0$, therefore $\|(x,0)\|=\max\{\|x\|_X,\|0\|_Y\}=\|x\|_X\gt 0$ (and the same argument can be made for $x=0$ and $y\neq 0$). Finally, let $x\neq 0$ and $y\neq 0$, then $\|x\|_X\gt 0$ and $\|y\|_Y\gt 0$ so $\|(x,y)\|\gt 0$. Therefore $\|(x,y)\|=0$ if and only if $(x,y)=(0,0)$.

2) Let $\alpha\in\mathbb{R}$ and $x\in X$ and $y\in Y$ be given. Then, as $\|x\|_X$ and $\|y\|_Y$ are norms:

$$\|(\alpha x,\alpha y)\|=\max\{\|\alpha x\|_X,\|\alpha y\|_Y\}=\max\{|\alpha|\|x\|_X,|\alpha|\|y\|_Y\}=|\alpha|max\{\|x\|_X,\|y\|_Y\}=|\alpha|\|(x,y)\|$$

3) Given $x_1,x_2\in X$ and $y_1,y_2\in Y$, as $\|x\|_X$ and $\|y\|_Y$ are norms:

$$\|(x_1+x_2,y_1+y_2)\|=\max\{\|x_1+x_2\|_X,\|y_1+y_2\|_Y\}\\\leq \max\{\|x_1\|_X+\|x_2\|_X,\|y_1+y_2\|_Y\}\\ \leq \max\{\|x_1\|_X+\|x_2\|_X,\|y_1\|_Y+\|y_2\|_Y\}\\ \leq \max\{\|x_1\|_X,\|y_1\|_Y\}+\max\{\|x_2\|_X,\|y\|_Y\}\\= \|(x_1,y_1)\|+\|(x_2,y_2)\|$$

Therefore $\|(x,y)\|$ is a norm on $X\times Y$

Banach space proof

Let $(x_n,y_n)$ be any Cauchy sequence in $X\times Y$, and as $x_n$ is Cauchy in $X$ and $y_n$ is Cauchy in $Y$, and $X$ and $Y$ are Banach spaces, $\exists x\in X, y\in Y$ such that $x_n\to x$ and $y_n\to y$. Then

$$\lim_{n\to\infty}\|(x_n,y_n)-(x,y)\|=\|\lim_{n\to\infty}(x_n,y_n)-(x,y)\|=\|(\lim_{n\to\infty} x_n,\lim_{n\to\infty} y_n)-(x,y)\|=\|(x,y)-(x,y)\|=max\{\|x-x\|_X,\|y-y\|_Y\}=max\{\|0\|_X,\|0\|_Y\}=max\{0,0\}=0$$

So $(x,y)\in X\times Y$, and therefore $X\times Y$ is complete in the norm, and therefore $X\times Y$ is a Banach space.

Answer 1

You need to show:

1. Let $(x_n,y_n)$ be any Cauchy sequence in $X\times Y$, then $(x_n)$ is Cauchy in X, and $(y_n)$ is Cauchy in Y.

If you can show 1, since $X,Y$ are Banach space, $\exists x\in X,y\in Y$, s.t. $x_n\to x,y_n\to y$.

2. Then you need to show $||(x_n,y_n)-(x,y)||\to 0$ in $X\times Y$.

Since $x_n\to x,y_n\to y$, $\forall \epsilon >0, \exists N, s.t. ||x_n-x||_X<\epsilon, ||y_n-y||_Y<\epsilon \quad \forall n\geq N$,

Then $\forall n\geq N, \quad ||(x_n,y_n)-(x,y)||=\max\{||x_n-x||,||y_n-y||\}<\epsilon$

Therefore $X\times Y$ is a Banach space.