Is the 2-Norm Always Less Than or Equal to the 1-Norm?

[fernanda2w5t83]

Homework Statement

Show that 2-norm is less equal to 1-norm

But I've found this proof

http://img825.imageshack.us/img825/5451/capturaklt.jpg

Which basically shows that if p=1 and q=2 then 1-norm is less equal than 2-norm, i.e. the opposite hypothesis

Homework Equations

None

The Attempt at a Solution

It's not homework, it's just a doubt

 

[jbunniii]

In a general measure space, it need not be true that ##\|x\|_p \leq \|x\|_q## or ##\|x\|_q \leq \|x\|_p##. Indeed, in general it is not even true that ##L^p \subset L^q## or ##L^q \subset L^p##. In order to obtain this nesting, it is necessary to impose additional restrictions.

If the space has finite measure (certainly the case for a probability measure space), then ##L^q \subset L^p## if ##1 \leq p \leq q \leq \infty##. The probability space is special because it has measure 1, which gives us ##||x||_p \leq ||x||_q##. In general, there would be a constant ##|x|_p \leq c \|x\|_q##. The intuitive explanation for why the containment goes this way is that a function in a finite measure space can have "thick singularities," but it can't have "thick tails."

If the space uses counting measure (or in general, if it does not contain sets with arbitrarily small positive measure), then ##L^p \subset L^q##. For example, this is true for the sequence spaces ##\ell^p \subset \ell^q##. The intuitive reason the containment goes in this direction is that you can have "thick tails" but not "thick singularities."

So it depends on what measure space you are working with. If we are in a finite-dimensional vector space, say ##\mathbb{R}^n##, then we have both ##\ell^p(\mathbb{R}^n) \subset \ell^q(\mathbb{R}^n)## and ##\ell^q(\mathbb{R}^n) \subset \ell^p(\mathbb{R}^n)##. In the special special case ##p = 1##, ##q = 2##, we will have ##\|x\|_2 \leq \|x\|_1## and ##\|x\|_1 \leq \sqrt{n} \|x\|_2##.