Let $(X,\| \cdot \|)$ be a normed linear space over $\Bbb K$. For $x \in X$ show that $$\|x\| = \inf \left \{ \frac 1 {|t|} : t \in \Bbb K \setminus \{0 \}, \|tx\| \leq1 \ \right \}.$$
Let $m=\inf \left \{ \frac 1 {|t|}: t \in \Bbb K \setminus \{0 \}, \|tx\| \leq1 \right \}.$ Then It is easy to see that $\|x\| \leq m$. But I find difficulty to prove the reverse direction. How do I proceed for proving the other way round? Please help me in this regard.
Thank you very much.
Let $A$ denote the set $\left\{ \frac 1 {|t|} : t \in \Bbb K \setminus \{0 \}, \|tx\|\leq 1 \ \right \}$, so that we have $m=\inf (A)$.
Assume first that $x$ is not zero. Note that if we consider $t=\frac{1}{\| x\| }\in \Bbb K \setminus \{0 \}$, then we have $$\|tx\|=\left \| \frac{x}{\| x\|} \right \|=\frac{\| x\|}{\| x\|}=1$$ by homogeneity of the norm. Hence, by definition, $\frac{1}{|t|}=\|x \| \in A$. It follows that $m\leq \|x \|$ by definition of the infimum of a set.
Assume now that $x=0$. Then $A=\left\{ \frac 1 {|t|} : t \in \Bbb K \setminus \{0 \}\right \}$, whose infimum can be seen to be $0$ by letting $|t|$ take arbitrary big values (it can be done by homogeneity: consider any fixed $t_1\in \mathbb{K}$ such that $|t_1|>0$, then define $t_n=n\cdot t_1$ for $n\in \mathbb{N}$, so that $|t_n|=n\cdot |t_1|$ goes to infinity).