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Let $X,Y$ be normed $\mathbb K$ linear space . $X \ne \{0\}$ and let $T$ be a $\mathbb K$ linear function. Then $$\|T\|= \sup\left\{\frac{\|T(x)\|}{\|x\|} : x \in X , x \ne 0 \right\}.$$ Here $\|T\| = \sup \{\|T(x)\|: x \in X, \|x\| \leq 1 \}$

How can I prove the inequality $$\|T\| \leq \sup \left\{\frac{\|T(x)\|}{\|x\|} :x \in X, x \neq 0\right\}$$ because the other inequality I have done.

Please someone give some hints.

Thank you..

Xander Henderson
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MADmind
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    Let $x \neq 0$ and $y = \frac1{|x|} x$. Then $|y|\le 1$. What can you say about $\frac{|T(y)|}{|y|}$? Note also that you can prove $|T| = \sup {\frac{|Tx|}{|x|}: |x| =1}$. – Cauchy Aug 17 '17 at 15:25

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