Consider an [math]{\mathcal L}^2[/math]-integrable function [math]f

>>7779799
let f(x)=ln|x| for |x|>1 f(x)=1/2*x^2 - 1/2 for |x|<1
∫ |f'|^2 is finite
∫ |f'| is infinite

>>7779799
>reading zizek
I'd rather by a slob drinking beer in front of a computer.

>>7779833
but the log isn't in L^2 in the first place

>>7779799

It's depressing that the artist's go-to author to illustrate the cultured, modern man, the first thing to pop into the artist's mind, is Zizek. The image would have been improved significantly if there were no text on the book's cover at all, and not lost any of its meaning.

>>7779799

ab <= a^2/2 + b^2/2

if a= f', b=1
then f' <= f'^2/2

>>7781434
You mean
f' <= f'^2/2+1/2
?

Don't see how it helps

>>7779799
A little more information would be helpfull. Do you mean integrable over [math]\mathbb{R}[/math] or over some bounded Subset? Is f differentiable/weakly differentiable/absolutley continuous? What about f'?

>>7781456
[eqn] f \in W^{1,2}( \mathbb{R}) [/eqn]

>>7781503
Sorry all the usal embenddings and inequalities (Sobolev, Rellich, Poincare-Wirtinger...) only work for bounded domoins as far as I know. But if anyone else has an answer i would definetly be interested.

>>7781545
The above wasn't me, but I'm mostly thinking about integration over all of R (as the function type implies)

I've googled the common inequalities but couldn't find anything

>>7779799
What about f(x)=x^{1/3} then f'(x)=x^{-2/3}/3 and |f'(x)|^2=|x^{-4/3}/3| which is integrable while f' isn't.

>>7781641
>f(x)=x^{1/3}
Not in L^2.