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Is it right to say that if two vectors, $A$ and $B$ (all elements of $A$ and $B$ are positive), have same $L^p$ norms, for all p, then $A = B$ ?. Thanks.

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    $\begingroup$ What do you think of the $l^p$ norms of $A$ and $-A$? $\endgroup$
    – Tryss
    Aug 28, 2015 at 14:44
  • $\begingroup$ Thanks. Tryss. Let me rephrase the question, and include (A /= -B). $\endgroup$
    – rex
    Aug 28, 2015 at 14:45
  • $\begingroup$ @rex: same issue with $(2,2,1,1,2)$ and $(-2,2,1,-1,2)$. Maybe you want to assume that $A$ and $B$ have non-negative components. $\endgroup$ Aug 28, 2015 at 14:53
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    $\begingroup$ Then don't we have $\|(1,2)\|_p = \|(2,1)\|_p$ for all norms $\| \cdot \|_p$ on $\mathbb R^2$? $\endgroup$
    – Simon S
    Aug 28, 2015 at 14:57
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    $\begingroup$ (By the way, writing "$L^p$ norms" and then talking about components is confusing. The $L^p$ spaces are well defined vector spaces of functions with a specific norm. Because you are talking about components, I take it you actually mean the $\| \cdot \|_p$ norm applied to the vector spaces $\mathbb R^n$.) $\endgroup$
    – Simon S
    Aug 28, 2015 at 15:02

3 Answers 3

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As Crostul and Simon S pointed out, there is always the possibility of a permutation of the coordinates which leaves the norm invariant.

But one can show that these are indeed the only such possibilities i.e. that if two increasing sequences $(a_i)_{i=1}^n, (b_i)_{i=1}^n$ with positive elements satisfy $$\sum_{i=1}^n a_i^p=\sum_{i=1}^n b_i^p$$ for all $p$ then you must have $a_i=b_i$ for all $i$.

This can be proved by an iterated argument using the fact that for $p \to \infty$ the $p$-norm of a vector assumes the value of the maximal coordinate. Hence we conclude that $a_n=b_n$ must hold.

Then, looking at the sequences $(a_i)_{i=1}^{n-1}$ and $(b_i)_{i=1}^{n-1}$, they must still satisfy the equation and hence we conclude $a_{n-1}=b_{n-1}$

Iterating this we obtain $a_i=b_i$ for all $i$. (Writing formally, we use mathematical induction here.)

So the possibilities of permutation mentioned above are indeed the only such possibilities.

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No. For a vector $v=(x_1, \dots, x_n)$ the $p$-norm is $$||v||_p = \left( \sum_i |x_i|^p \right)^{\frac1p}$$ so you can see that for any permutation of indexes $\sigma \in S_n$ you have $$||v||_p = ||(x_{\sigma(1)}, \dots, x_{\sigma(n)})||_p$$

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Take A=(1,1,2) and B=(1,2,-1) and A is not equal to B For p=2 , N(A)=sqrt(1^2+1^2+2^2)=sqrt(6) N(B)=sqrt(1^2+2^2+1^2)=sqrt(6) i.e N(A)=N(B) But A is not equal to B Therefore This result is not true in general

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