# functional analysis – Equivalent norms in the intersection

Let $$V$$ be a vector space. Two norms $$|cdot |_1,|cdot |_2: V longrightarrow mathbb{R}$$ in $$V$$ are said equivalent if there exists $$a,b>0$$ such that
$$a|u|_1 leq |u|_2 leq b |u|_1,; forall ; u in V.$$

Now,, consider $$X=(X, |cdot|_X)$$ and $$Y=(Y, |cdot|_Y)$$ two normed spaces. It is easy to show that $$Z:= X cap Y$$ is a normed space with norm $$|cdot|_Z: Z longrightarrow mathbb{R}$$ given by
$$|u|_Z=|u|_X+|u|_Y,; forall ; u in Z.$$

Question. If, for some some $$alpha, beta>0$$, we consider $${rm N}: Z longrightarrow mathbb{R}$$ as
$${rm N}(u)=alpha |u|_X +beta |u|_Y,; forall ; u in Z$$
then $${rm N}$$ defines a equivalent norm (in $$Z$$) with respect the norm $$|cdot|_Z$$?

I don’t see, for instance, how to prove that
$$alpha |u|_X +beta |u|_Y leq gamma( |u|_X + |u|_Y),; forall ; u in Z tag{1}$$
for some $$gamma>0$$. Can I to use some property of $$max$$ or $$min$$? Or, can I only prove $$(2)$$ instead of the equivalence of norms? The equivalence is more stronger than $$(2)$$.