If $(X_1,||.||_1)$ and $(X_2,||.||_2)$ are two normed spaces and define norm on $X_1\times X_2$ as $||x||=\max(||x_1||_1,||x_2||_2)$. I want to check the triangle inequality property for this norm, i.e, $||x+y||\leq||x||+||y||$. After using using triangle inequality for the two norms I get $||x+y||\leq\max(||x_1||_1+||y_1||_1,||x_2||_2+||y_2||_2)$. My question is why is $\max(||x_1||_1+||y_1||_1,||x_2||_2+||y_2||_2)\leq\max(||x_1||_1,||x_2||_2)+\max(||y_1||_1,||y_2||_2)$
[Math] norm of product of normed spaces
functional-analysis
Related Solutions
(I). We want to show that for any $(p_1,p_2)\in U_1\times U_2$ there exists $r_3>0$ such that the open $d_+$ ball $B_{d_+}((p_1,p_2),r_3)$ is a subset of $U_1\times U_2.$
Take $r_1,r_2>0$ such that $B_{d_1}(p_1,r_1)\subset U_1$ and $B_{d_2}(p_2,r_2)\subset U_2.$ Let $r_3=\min (r_1,r_2).$ Then $$(p,p')\in B_{d_+}((p_1,p_2),r_3)\implies$$ $$\implies (\;d_1(p,p_1)<r_3\leq r_1\;\land\; d_2(p',p_2)<r_3\leq r_2\;) \implies$$ $$\implies (\; p\in B_{d_1}(p_1,r_1)\subset U_1 \;\land \;p'\in B_{d_2}(p_2,r_2)\subset U_2\;)\implies$$ $$\implies (p,p')\in U_1\times U_2.$$
Since this holds for all $(p,p')\in B_{d_+}((p_1,p_2),r_3),$ we have $B_{d_+}((p_1,p_2),r_3)\subset U_1\times U_2. $
(II). Metrics $d$ and $d'$ on a set $X$ produce the same topology iff $$(i).\quad \forall p\in X \;\forall r>0 \;\exists s>0 \;(\;B_d(p,s)\subset B_{d'}(p,r)\;),\text { and }$$ $$ (ii). \quad \forall p\in X \;\forall r>0 \;\exists s'>0\;(\;B_{d'}(p,s')\subset B_d(p,r)\;).$$
To get an idea of how to use this to show that $d_e$ and $d_{max}$ generate the same topology as $d_+,$ consider the case $X_1=X_2=\mathbb R,$ with $d_1(x,y)=d_2(x,y)=|x-y|.$ Sketch some pictures of open balls of various radii, centered at some $p\in \mathbb R^2$, with respect to these 3 metrics.
Metrics on a set $X$ that generate the same topology on $X$ are called equivalent metrics.
This is true for arbitrary metric spaces of which normed spaces are just a special case.
Given metric spaces $(X_1, d_1), \ldots, (X_n, d_n)$ the product topology on $X = X_1 \times \cdots \times X_n$ is generated by the metric $$ d(x, y) = \max\{d_1(x_1, y_1), \ldots , d_n(x_n, y_n)\} $$ You can verify that this is indeed a metric. Also for normed spaces $d_k(x_k, y_k)$ is just $||x_k - y_k||_k$.
For let $U = U_1 \times \cdots \times U_n$ be a basis element of the product space $X$ where each $U_k$ is open in $X_k$. And let $x = (x_1, \ldots, x_n)$ be a point in $U$. Then you can show that there is an open $d$-ball around $x$ contained inside $U$.
Indeed note that as each $U_k$ is open there are open $d_k$-balls $B_{d_k}(x_k, r_k) \subseteq U_k$ around each component $x_k$ in $X_k$. Now let $r = \min\{r_1, \ldots, r_n\}$. Then you can verify that $B_d(x, r)$ is an open $d$-ball around $x$ in $X$ such that $B_d(x, r) \subseteq U$.
Conversely let $B_d(x, r)$ be an open $d$-ball around $x$ in $X$. You can show that there is a basis element $U = U_1 \times \cdots \times U_n$ around $x$ that is contained inside that open ball.
Indeed just let $U = B_{d_1}(x_1, r) \times \cdots \times B_{d_n}(x_n, r)$ and we are done.
Hence every open set with respect to the product topology on $X$ is also an open set with respect to the above $d$-metric on $X$. And every open set with respect to the same $d$-metric on $X$ is also an open set with respect to the product topology on $X$ as expected.
Best Answer
$$\lVert x_1\rVert_1+\lVert y_1\rVert_1\le \max\{\lVert x_1\rVert_1,\lVert x_2\rVert_2\}+\max\{\lVert y_1\rVert_1,\lVert y_2\rVert_2\}$$ $$\lVert x_2\rVert_2+\lVert y_2\rVert_2\le \max\{\lVert x_1\rVert_1,\lVert x_2\rVert_2\}+\max\{\lVert y_1\rVert_1,\lVert y_2\rVert_2\}$$