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.):


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: $<v, v>=0 \Rightarrow v=0$ 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:

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

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.

  • 2
    $\begingroup$ 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. $\endgroup$ – DonAntonio May 31 at 13:57

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