Comparison of two norms.

Question

Let $\| \cdot \|_{1}, \| \cdot \|_{2}$ be two norms on a $\Bbb K$-linear space $X$. Suppose $\| x \|_{1} < 1 \implies \|x\|_{2} < 1$. Prove that $\|x \|_{2} \leq \|x\|_{1}$ for all $x \in X$.

I know that if $\| \cdot \|$ is a norm on $X$ then $$\|x\| = \inf \left \{ \frac 1 {|t|} : t \in \Bbb K \setminus \{0 \}, \|tx\| \leq1 \ \right \}.$$ Now let us consider sets $A$ and $B$ such that $A=\left \{ \frac 1 {|t|} : t \in \Bbb K \setminus \{0 \}, \|tx\|_{1} \leq1 \ \right \}$ and $B = \left \{ \frac 1 {|t|} : t \in \Bbb K \setminus \{0 \}, \|tx\|_{2} \leq1 \ \right \}.$ I have to show that $A \subset B$. For this let us take $y \in A$. Then $\exists$ $t \in \Bbb K \setminus \{0 \}$ with $\|tx\|_{1} \leq 1$ such that $y = \frac 1 {|t|}$. If $\|tx\|_{1} < 1$ then by the given hypothesis it follows that $\|tx\|_{2} < 1$. But that means $y = \frac 1 {|t|} \in B$, which fulfills our requirement. But I find difficulty when $\|tx\|_{1}=1$. Please help me in this regard.

Thank you very much.

Answer 1

Suppose there exists $x$ such that $\|x\|_2>\|x\|_1$, there exists $t>0$ such that $t\|x\|_1=\|tx\|_1<1$ and $\|tx\|_2\geq 1$, for example $t={1\over{\|x\|_2}}$contradiction.

Answer 2

Here is an approach that avoids all that Minkowski mess.

The statement is equivalent to $B_1(0,1) \subset B_2(0,1)$ and hence $B_1(0,r) \subset B_2(0,r)$ for all $r>0$.

Note that $\|x\|_* = \inf_{r>0} \{ r | \|x\|_* < r \} = \inf_{r>0} \{ r | x \in B_*(0,r) \}$.

Finally, since $ \{ r | x \in B_1(0,r) \} \subset \{ r | x \in B_2(0,r) \}$ we have $\|x\|_2 = \inf_{r>0} \{ r | x \in B_2(0,r) \} \le \inf_{r>0} \{ r | x \in B_1(0,r) \} = \|x\|_1$.