# Cauchy Schwarz inequality using L1 norm

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 $|A^Tb|\leq\Vert A\Vert_F \Vert b\Vert_2$

• 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 $n^{2/p}$. Jul 14, 2015 at 5:37
• @Jose27 can you comment on the first answer, as i thought the same answer you suggested. Jul 14, 2015 at 5:48
• 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. Jul 14, 2015 at 8:23
• @Chee Han So does that mean inequality using L1 holds true. And what about the second inequality i asked for. Jul 14, 2015 at 11:31
• 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) Jul 14, 2015 at 13:51

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 $\int fg =+\infty$.
• I think OP meant the setting to be $\mathbb{R}^n$ (i.e. $\{ 1,\ldots, n\}$ with counting measure), not a general measure space. Jul 14, 2015 at 19:49