1
$\begingroup$

1) Cauchy-Schwarz inequality states that the absolute value of vector inner product is always less or equal to product of norms of individual vectors i.e., $|a^Tb|\leq\Vert a\Vert_2 \Vert b\Vert_2$.

Does this inequality hold true for any Lp vector norms especially the L1 norm i.e., $|a^Tb|\leq\Vert a\Vert_1 \Vert b\Vert_1$.? If yes any Proofs?


2) Does this Cauchy-Schwarz inequality hold for a matrix $A$ having linearly independent columns $|A^Tb|\leq\Vert A\Vert_F \Vert b\Vert_2$

$\endgroup$
7
  • $\begingroup$ It's true for any $p$-norm with $1\leq p \leq 2$ and false for $p>2$. To see this simply note that $\| \cdot \|_2\leq \| \cdot \|_p$ whenever the first condition is satisfied. On the other hand if $p>2$ then with $a=b=(1,\ldots,1)$ the LHS is $n$ (the dimension), while the RHS is $n^{2/p}$. $\endgroup$
    – Jose27
    Jul 14, 2015 at 5:37
  • $\begingroup$ @Jose27 can you comment on the first answer, as i thought the same answer you suggested. $\endgroup$
    – Astro
    Jul 14, 2015 at 5:48
  • 1
    $\begingroup$ Here is the reason why: Cauchy-Schwarz inequality holds true for vectors in an inner product space; now inner product gives rise to a norm, but the converse is false. $\endgroup$
    – Chee Han
    Jul 14, 2015 at 8:23
  • $\begingroup$ @Chee Han So does that mean inequality using L1 holds true. And what about the second inequality i asked for. $\endgroup$
    – Astro
    Jul 14, 2015 at 11:31
  • 1
    $\begingroup$ The $L^1$ norm doesn't satisfy the parallelogram identity, so it is not induced by an inner product, hence the inequality fails. Note that this is a necessary and sufficient condition. Since all norms on finite dimensional space are equivalent, I suspect the second inequality is true. (possibly with a constant) $\endgroup$
    – Chee Han
    Jul 14, 2015 at 13:51

1 Answer 1

2
$\begingroup$

No it doesn't hold in $L^1$. Take $f(x)=g(x)=\frac{1}{\sqrt{x}}$ for $x \in (0,1)$ and $f(x)=g(x)=0$ elsewhere.

$\Vert f \Vert_1=\Vert g \Vert_1=2$ but $\int fg =+\infty$.

$\endgroup$
3
  • $\begingroup$ Thanks for help, but can you explain in terms of vectors with transpose involved. And what about second part.? $\endgroup$
    – Astro
    Jul 14, 2015 at 5:43
  • $\begingroup$ I think holder's inequality which is valid for all Lp norms, for p=q=2 is same as Cauchy Schwarz inequality. $\endgroup$
    – Astro
    Jul 14, 2015 at 6:17
  • 1
    $\begingroup$ I think OP meant the setting to be $\mathbb{R}^n$ (i.e. $\{ 1,\ldots, n\}$ with counting measure), not a general measure space. $\endgroup$
    – Jose27
    Jul 14, 2015 at 19:49

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.