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


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.


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).

  • $\begingroup$ 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 '19 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 '19 at 20:11
  • $\begingroup$ @mhj I mean to say that I've not talking about the transformation that you mentioned. $\endgroup$ – cmk Jun 3 '19 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 '19 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 '19 at 20:39

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.