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For $(\mathbb{R}^2,\|\cdot\|_2)$ and $(\mathbb{R}^2,\|\cdot\|_\infty)$ and any $x \in B((0,0),1,\|\cdot\|_2)$ how would you find a $\delta_x$ such that $B(x,\delta_x,\|\cdot\|_\infty) \subset B((0,0),1,\|\cdot\|_2)$ ?

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Think about this picture and this one –  t.b. Apr 7 '12 at 11:34

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Geometrically $B(x,\delta_x,\|\cdot\|_\infty)$ is a box of side length $2\delta_x$. The furthest a point in the box can be (in the Euclidean metric) from the point $x$ is at one of the corners, which is $\sqrt{2}\delta_x$ away from the midpoint. Thus we have an inclusion of balls $B(x,\delta_x,\|\cdot\|_\infty)\subseteq B(x,\sqrt{2}\delta_x,\|\cdot\|_2)$. Now the Euclidean distance from $x$ to the edge of the unit $\|\cdot\|_2$-ball is $1-\|x\|_2$, so we also have

$$B\left(x,t(1-\|x\|_2),\|\cdot\|_2\right)\subset B(O,1,\|\cdot\|_2)$$

for any $0<t<1$. Choosing such a $t$, we may set

$$\delta_x:=\frac{t(1-\|x\|_2)}{\sqrt{2}}$$

so that we ultimately have

$$B(x,\delta_x,\|\cdot\|_\infty)\subseteq B(x,\sqrt{2}\delta_x,\|\cdot\|_2)=B\left(x,t(1-\|x\|_2),\|\cdot\|_2\right)\subset B(O,1,\|\cdot\|_2).$$

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Since for all $v\in\mathbb R^2$ we have $\lVert v\rVert_2\leq \sqrt 2\lVert v\rVert_{\infty}$, we have for all $x$ and $r>0$ that $B(x,\sqrt 2r,\lVert \cdot\rVert_{\infty})\subset B(x,r,\lVert \cdot\rVert_2)$, so we just need to find $r_x$ such that $B(x,r_x,\lVert \cdot\rVert_2)\subset B((0,0),1,\lVert\cdot\rVert_2)$ (then we will take $\delta_x=\sqrt 2r_x$). Check that $r_x=\frac{1-\lVert x}2$ works.

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