# normed space - inner product

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:

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

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.

## 1 Answer

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).

• 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) ? – mhj Jun 3 '19 at 20:04
• 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. – cmk Jun 3 '19 at 20:11
• @mhj I mean to say that I've not talking about the transformation that you mentioned. – cmk Jun 3 '19 at 20:33
• 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? – mhj Jun 3 '19 at 20:37
• 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! – cmk Jun 3 '19 at 20:39