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I'm stuck with the following exercise. Hopefully some of you can help me.

Exercise

A normed K-vectorspace is a $\mathbb{K}$ -vectorspace $V,$ where a transformation is defined as:

$\|\cdot\| : V \rightarrow[0, \infty[$

for all $v, w \in V$, where the following conditions has to be met:

a) $\|v\|=0$ if and $\mathrm{only}$ if $v=0 .$
b) $\|\alpha \cdot v\|=|\alpha| \cdot\|v\|$ for $\alpha \in \mathbb{K}$
c) $\|v+w\| \leq\|v\|+\|w\|$

1) Show, that the norm on a inner product space $V$ makes $V$ to a normed vectorspace.

In the following questions is considered the real vectorspace $V=\mathbb{R}^{2}$ and the following transformation:
$$\|\cdot\| : \mathbb{R}^{2} \rightarrow[0, \infty[$$ $$\quad \left( \begin{array}{l}{\alpha} \\ {\beta}\end{array}\right) \mapsto|\alpha|+|\beta|$$

We define:
$\langle\boldsymbol{v}, \boldsymbol{w}\rangle=\frac{1}{4}\left(\|\boldsymbol{v}+\boldsymbol{w}\|^{2}-\|\boldsymbol{v}-\boldsymbol{w}\|^{2}\right)$ for all $\boldsymbol{v}, \boldsymbol{w} \in \mathbb{R}^{2}$

5) Conclude, that the transformation $\|\cdot\|$ above, is not defined from the inner product on $V=\mathbb{R}^{2}($ Hint: Polarization identity $)$

My approach

1) What am I supposed to do here?

5) Here is my approach:

$|\alpha|+|\beta|=\|v\|=(<v, v>)^{\frac{1}{2}}=\left(\frac{1}{4}\left(\|v+v\|^{2}-\|v-v\|^{2}\right)\right)^{\frac{1}{2}}=\left(\frac{1}{4}\|2 v\|^{2}\right)^{\frac{1}{2}}=\left(\|v\|^{2}\right)^{\frac{1}{2}}=\|v\|$

So we can see that the transformation is defined by $\langle\boldsymbol{v}, \boldsymbol{w}\rangle$, but we just showed above in 4) that $\langle\boldsymbol{v}, \boldsymbol{w}\rangle$ is not an inner product.

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I will answer your first question:

Every inner product $\langle \cdot,\cdot\rangle$ induces a norm $\left\lVert\cdot\right\rVert$ via $$\left\lVert x\right\rVert=\sqrt{\langle x,x\rangle}.$$ They want you to prove that this is, indeed a norm i.e. that the above function satisfies the norm properties.

I'm not sure what your question is for 5).

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  • $\begingroup$ But the transformation: $$\|\cdot\| : \mathbb{R}^{2} \rightarrow[0, \infty[$$ $$\quad \left( \begin{array}{l}{\alpha} \\ {\beta}\end{array}\right) \mapsto|\alpha|+|\beta|$$ doesn't belong to question 1). So should I just show that the norm indeed meets the conditions a),b),c) ? $\endgroup$ – mhj Jun 3 at 20:04
  • $\begingroup$ Right, I'm not talking about the function. I'm talking about the function $\left\lVert x\right\rVert=\sqrt{\langle x,x\rangle}.$ You want to show that this constitutes a norm. $\endgroup$ – cmk Jun 3 at 20:11
  • $\begingroup$ @mhj I mean to say that I've not talking about the transformation that you mentioned. $\endgroup$ – cmk Jun 3 at 20:33
  • $\begingroup$ Ahh okay, so like this? The norm is by definition: $\|v\|=\sqrt{\langle v, v\rangle}$ and the definition of a inner product implies: $<v, v>=0 \Rightarrow v=0$ Which implies $\|v\|=0 $ if and only if $ v=0$ <br> - But I mean, that's just the defintion of a norm? $\endgroup$ – mhj Jun 3 at 20:37
  • $\begingroup$ Yes, what you're suppose to show is that $\sqrt{(v,v)}$ is a norm, and you're going to use lots of inner product properties to get there. So, your work here shows that $\sqrt{(v,v)}$ satisfies the first property! $\endgroup$ – cmk Jun 3 at 20:39

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