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An inner product space $E$ becomes a normed linear space when equipped with the norm $\|x\|=\sqrt{\langle x,x\rangle}\,\,$ for all $x \in E$

I tried proving the above theorem myself by using the three properties of inner product to prove the three axioms/properties of normed linear spaces. I was able to establish the following:

N1: $\|x\|\geq0$

Proof: $$\langle x,x\rangle\geq0$$ by definition Therefore $$\|x\|=\sqrt{\langle x,x\rangle}\,\,\geq0$$

N2: $\|\lambda x\|=|\lambda|\|x\|$

Proof: $$\langle\lambda x, \lambda x\rangle=\lambda\langle x, \lambda x\rangle$$ This was where I got stuck. Can I simplify further to this? $$\lambda\langle x, \lambda x\rangle={\lambda}^2 \langle x, x\rangle$$ If I can, then I can proceed further. If not, please I need guidance.

N3: $\|x+y\|\leq\|x\|+\|y\|$

Proof:

I started with the relation: $$\|x+y\|=\sqrt{\langle x+y,x+y\rangle}$$ Thus $${\|x+y\|}^2=\langle x+y,x+y\rangle=\langle x, x+y\rangle+\langle y, x+y\rangle=\langle x, x\rangle+\langle x, y\rangle+\langle y, x\rangle+\langle y, y\rangle$$ This simplifies further to: $${\|x+y\|}^2={\|x\|}^2+\langle x, y\rangle+\bar{\langle y, x\rangle}+{\|x\|}^2$$ I got confused here too. How is $\langle x, y\rangle,+\bar{\langle y, x\rangle}=2\Re \langle x, y\rangle\leq 2\langle x, y\rangle $

I know this might seem trivial but I'd appreciate if someone can help me out.

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For condition 2, we have $$ \langle \lambda x,\lambda x\rangle =\lambda^2\langle x,x\rangle $$ if the vector space is real, and $$ \langle \lambda x,\lambda x\rangle =|\lambda|^2\langle x,x\rangle $$ if it is complex. In both cases we get $||\lambda x||=|\lambda|\cdot||x||$ after taking a square root.

To prove condition 3, you will need the Cauchy-Schwarz inequality $$ |\langle x,y\rangle|\leq ||x||||y||$$ together with the fact that $\Re z\leq |z|$ for all complex numbers $z$.

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  • $\begingroup$ Thanks carmichael561, your response were apt $\endgroup$ – Obinoscopy Oct 5 '16 at 19:14

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