# Inner product space - normed space

I'm kind off stuck with the following exercise. Hopefully some of you can help me (It's mainly question 1 and 5, Im uncertain about. You don't need to help me with 3 and 4.):

## 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:
$$\|\cdot\| : \mathbb{R}^{2} \rightarrow[0, \infty[$$ $$\quad \left( \begin{array}{l}{\alpha} \\ {\beta}\end{array}\right) \mapsto|\alpha|+|\beta|$$ 2) Show, that $$\mathbb{R}^{2}$$ and $$\|\cdot\|$$ makes a normed vectorspace.

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 $$\boldsymbol{v}, \boldsymbol{w} \in \mathbb{R}^{2}$$

3) Show, that $$(1),(2)$$ og $$(3)$$ in Definition 9.1 is met for $$\langle\cdot, \cdot\rangle$$

4) Show, that property $$(4)$$ in Definition 9.1 is not met for $$\langle\cdot,\rangle$$ (Hint: consider the vectors $$\boldsymbol{u}=\boldsymbol{v}=\left( \begin{array}{l}{1} \\ {1}\end{array}\right), \boldsymbol{w}=\left( \begin{array}{l}{2} \\ {0}\end{array}\right)$$ and scalars $$\alpha=\beta=1 )$$

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? Do I just need to show that the definition of a inner product is compatible with definition a),b),c) for a normed vector space?

a) $$\|v\|=\sqrt{\langle v, v\rangle}$$ Where the definition for a inner product implies: $$=0 \Rightarrow v=0$$ Which implies $$\|v\|=0 \Rightarrow v=0$$
b) $$\|\alpha v\|=\sqrt{\langle\alpha v, \alpha v\rangle}=\sqrt{\alpha{\overline{\alpha}} \langle v, v \rangle}==\sqrt{|\alpha|^{2}\langle v, v \rangle}=|\alpha|\|v\|$$
c) I'm not going to show this rn, since I believe I'm on the wrong track.

2) What am I supposed to do here, and how does it differ from 1)?
Is it just so simple, so taking the norm of an element is defined as above. So I should just show that taking the norm like above meets the definitions of a normed vector space?

Let $$v, v \in \mathbb{R}^{2},$$ where $$v=\left( \begin{array}{c}{\alpha} \\ {\beta}\end{array}\right)$$ and $$u=\left( \begin{array}{c}{a} \\ {b}\end{array}\right)$$

a) $$\|v\|=|\alpha|+|\beta|=0 \Rightarrow \alpha=\beta=0 \Rightarrow V=0$$

b) $$\|\gamma v\|=|\gamma \alpha|+|\gamma \beta|=|\gamma|(|\alpha|+|\beta|)=|\gamma|\|v\|$$

c) 2) $$\|v+u\|=\| \left( \begin{array}{c}{\alpha+a} \\ {\beta+b}\end{array}\right)\|=|\alpha+a|+|\beta+b|=(|\alpha|+|\beta|)+(|a|+|b|)=\| v\|+\| u \|$$

3) I need to show: (a) The scalar $$\langle v, v\rangle$$ is a real number and $$\langle v, v\rangle \geq 0$$ .
(b) $$\langle v, v\rangle= 0 \Rightarrow v=0$$
(c) $$\langle v, w\rangle=\langle w, v\rangle$$

For the defined inner product above. This is easily done, so you don't need to show me how this is done.

4) I need to show:
$$(\mathrm{d})\langle\alpha \cdot \boldsymbol{u}+\beta \cdot \boldsymbol{v}, \boldsymbol{w}\rangle \neq\alpha \cdot\langle\boldsymbol{u}, \boldsymbol{w}\rangle+\beta \cdot\langle\boldsymbol{v}, \boldsymbol{w}\rangle$$

Using the hint. The left hand side equal to 8 and right hand side equal to 6. So they're not equal.

5) I'm not sure how to show this. 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.

• An advise: don't write questions this long! It is almost certain that most people won't even read all that. Try thus to split the questions into sub-questions, and do post maybe 2-3 different questions. – DonAntonio May 31 at 13:57