Sorry, I can't access the image on my browser; it won't load for some reason. Could you please type out the inequality in Latex?

So I'll just have to guess what you might want; apologies if this is not what you're asking. The classical one-dimensional Hardy inequality

$$
\int_0^\infty \left| \frac{1}{t} \int_0^t f(s) ds \right|^2 dt \leq
4 \int_0^\infty |f(s)|^2 ds
$$

has a generalisation for weights: for any functions $m,w \geq 0$, the best constant $M=M(m,w)$ in the inequality

$$
\int_0^\infty \left| \int_0^t f(s) ds \right|^2 m(t) dt \leq
M \int_0^\infty |f(s)|^2 w(s)ds
$$

is related to the quantity
$$
S(m,w) = \sup_{R>0} \left( \int_R^\infty m(t) dt \right) \left( \int_0^R w(s)^{-1} ds \right)
$$
by $S \leq M \leq 4S$. Also, if either $M$ or $S$ equals $+\infty$, then so does the other (i.e. we don't have any reasonable Hardy inequality).

There are generalisations to $L^p$, $L^q$ norms, and $d\mu(t)$ a measure instead of just $m(t)dt$. If you look up Hardy's Inequality with Weights by B. Muckenhoupt from about 1972, and newer related papers, you'll get loads of articles.

However, the multidimensional questions are more difficult and much less is known.

So, if you just wanted a one-dimensional version on $[0,L]$ instead of $[0,\infty)$, you can just set $m(t) = 0$ and $w(s) = +\infty$ for $t,s > L$.