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Reading a script i've found task in which i had to determine whether each norm $$\|x\|_{p}=\left(\sum |x_{i}|^{p}\right)^{1/p}$$ origins from scalar product. Assuming $$p=2$$ i got, it comes from standard scalar product. $$\|x\|_{2}=\langle x,x\rangle^{1/2}$$ no luck for others. Tip i got was to use Parallelogram law and describe it in norm language. I get. $$\|x+y\|^{2}+\|x-y\|^{2}=2\|x\|^2+2\|y\|^{2}$$ I just don't see how should i use it.

vanHohenheim
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3 Answers3

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You can check that

$$x = (1, 0, 0, \cdots), \ \ \ y = (0,1,0, 0, \cdots)$$

in $\ell _p$ satisfies the parallelogram law only when $p = 2$. Indeed, the right hand side is $2\cdot 2^{\frac 2p}$ and the left hand side is $4$. Thus $\ell_p$-norm does not come from an inner product.

  • I'm probably missing something, but the example seems to satisfy the parallelogram law for $p=1$ as well. But then I think that $p$ must be even to ensure the norm is positive definite ? – Tom Collinge Oct 04 '15 at 15:00
  • I get $2 \cdot 2^{\frac 2p}$ on one side and $4$ the other side..@TomCollinge (Indeed now my answer is almost the same as that of Emilio) –  Oct 04 '15 at 15:05
  • With $p=1$ I get $(1+1)^2 +(1-1)^2 =4 = 2.1 + 2.1$ – Tom Collinge Oct 04 '15 at 15:50
  • See my extended comment posted as an answer. – Tom Collinge Oct 04 '15 at 18:55
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Let: $$ x=(1,1,0,0 \cdots) \qquad y=(1,-1,0,0 \cdots) $$ we have: $$ \|x\|_p=2^{1/p}=\|y\|_p $$ and $$ x+y=(2,0,0\cdots) \Rightarrow \|x+y\|_p=2 $$ $$ x-y=(0,2,0\cdots) \Rightarrow \|x-y\|_p=2 $$ so the parallelogram low becomes: $$ \|x+y\|_p^2+\|x-y\|_p^2=2\left(\|x\|_p^2+\|y\|_p^2 \right) $$ i.e: $$ 8=2\left(2^{2/p}+2^{2/p} \right) \iff 2=2^{2/p} $$ that gives $p=2$.

This means that the parallelogram low is verified only if $p=2$ and only in this case the norm can be derived from an inner product.

Tom Collinge
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Emilio Novati
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Consider this an extended comment......

Are you sure that your norm is $\|x\|_p=\left(\sum x_i^{p}\right)^{1/p}$ and not in fact $\|x\|_p=\left(\sum |x_i|^{p}\right)^{1/p}$

See for example the wiki entry on p-norm https://en.wikipedia.org/wiki/Norm_(mathematics)#p-norm.

If the two answers given assume the p-norm then they are correct.

If your original statement is correct, then it cannot be a norm for any odd value of $p$ since $\|(-1, 0, 0, ...)\|$ is not positive for $p=1$ and is not real for any odd $p > 1$.

Tom Collinge
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