3
$\begingroup$

Show whether $F: \mathbb{R}^3 \rightarrow \mathbb{R}$ defines a normed vector space on $\mathbb{R}^3$.

For $(\underline{x}) = (x_1, x_2, x_3)$,

$F(\underline{x}) = (\sum_{i=1}^{3} |x_i|^\frac{1}{2})^2$.

So at first glance this seems similar to the Euclidean norm, only with the powers swapped over. However I'm fairly sure this contradicts the triangle inequality for norms but I'm not sure how to show this.

$\endgroup$
  • 3
    $\begingroup$ Well, you just need a counterexample. Think of $a=(1,0,0)$ and $b=(0,1,0)$. Then $F(a+b) = 4 > 2 =F(a) + F(b) $. $\endgroup$ – Peter Oct 10 '15 at 14:16
  • $\begingroup$ How do you prove the p-norm is not a norm in $\mathbb{R}^n$ when 0<p<1 $\endgroup$ – Weaam Oct 10 '15 at 14:17
  • $\begingroup$ Ahh thank you so much! I tried a counter example but forgot to square at some stage. $\endgroup$ – C.B. Oct 10 '15 at 14:24
1
$\begingroup$

There's probably an even simpler example you can use, but this one works.

Consider the three points $(0,0,0)$, $(1,0,0)$ and $(2,2,2)$. Attempt to find the distances between them and apply the triangle inequality, and you should get something like $$1 + \left(1+2\sqrt{2}\right)^2\geq 18$$

or approximately

$$ 15.6 \geq 18$$

which is ridiculous.

$\endgroup$
  • $\begingroup$ Peter's example in the comments above is much simpler. $\endgroup$ – Zubin Mukerjee Oct 10 '15 at 14:20

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.