Is there an inequality more general thatn Hölder inequality? It is well known that for two given functions $f,g:\mathbb{R}^d \rightarrow \mathbb{R}^d$ such that $fg \in L^1(\mathbb{R}^d)$ and $f\in L^p(\mathbb{R}^d)$ and $g\in L^q(\mathbb{R}^d)$ with $\frac{1}{p}+\frac{1}{q}=1$, $p,q\geq 1$ then
$$\left|\int_{\mathbb{R}^d} f(x)g(x)dx\right|\leq \int_{\mathbb{R}^d} |f(x)g(x)|dx \leq \left(\int_{\mathbb{R}^d} |f(x)|^p\right)^{1/p} \left(\int_{\mathbb{R}^d}|g(x)|^q\right)^{1/q},$$
the latter known as Hölder inequality. This equality also holds when $"p=\infty"$ taking supremum.
$\bullet$ My question is: are there other type of inequalities that allow to "split" $f$ and $g$? Whatever inequality, more general or not, that allows to split functions $f$ and $g$. Any ideas? How or where could I find this information? Preferably one of the functions should be as general as possible, e.g. non-continuous.
Thanks a lot! :)
 A: You can check the following inequality. 
The well known young inequality claim that for every $a,b\in\mathbb{R}$
$$ |ab|\le \frac{a^2}{2} +\frac{b^2}{2}$$
taking $a= \sqrt{2\varepsilon}a'$ and $\frac{b'}{\sqrt{2\varepsilon}}$ one gets,
$$ |a'b'|\le \varepsilon a'^2+\frac{b'^2}{4\varepsilon}$$
Whence this leads to,
$$ \int_{\mathbb{R}^d} f(x)g(x) \, dx \le \Big(\varepsilon\int_{\mathbb{R}^d} f(x)^2 \, dx\Big) +\frac{1}{4\varepsilon}\Big(\int_{\mathbb{R}^d} g(x)^2 \, dx \Big) $$
that is 
$$\|fg\|_1\le \varepsilon\|f\|^2_2+\frac{1}{4\varepsilon}\|g\|^2_2\tag{1}$$
which holds for every $\varepsilon>$. The above inequality is more general than the well known Cauchy-schwartz inequality since one can recover it by taking $$ \varepsilon  =  \frac{\|g\|}{2\|f\|}~~~\text{when } ~~~~\|f\|\neq0.$$
For the general case, knowing that,
$$ |ab|\le \frac{a^p}{p} +\frac{b^q}{q}~~~\frac{1}{p}+\frac{1}{q}=1$$
by taking $ a= \sqrt[p]{p\varepsilon} a'$ and $ b= \frac{1}{\sqrt[p]{p\varepsilon} }b'$ we have 
$$ |a'b'|\le \varepsilon a'^p+\frac{b'^q}{q(\varepsilon p)^{\frac{1}{p-1}}}$$
therefore,
$$\|fg\|_1\le \varepsilon\|f\|^p_p +\frac{1}{q(\varepsilon p)^{\frac{1}{p-1}}}\|g\|^q_q\tag{2}$$
This last inequality doesn't necessary show that Holder is a special. Whereas, solving for $\varepsilon$
$$ \varepsilon\|f\|^p_p =\frac{1}{q(\varepsilon p)^{\frac{1}{p-1}}}\|g\|^q_q$$
Then, by carefully made use of the relationship $q= \frac{p}{p-1}$ we get 
$$ \varepsilon =\frac{1}{q^{\frac{1}{q}}p^{\frac{1}{p}}}\frac{\|g\|_q}{\|f\|^{p-1}_p}. $$
with this particular value of $\varepsilon $ in (2) we recover the following inequality. 
$$\|fg\|_1\le+\frac{2}{q^{\frac{1}{q}}p^{\frac{1}{p}}} \|f\|_p \|g\|_q\tag{3}$$
But (3) look like Holder inequality and (2) splits "$fg$" by integration and give (3) as special case.
this might help.
A: A special case of Hölder inequality is Schwarz's inequality:
$$\left(\int_{\mathbb R^d}f(x)g(x)\,dx\right)^2 \le \int_{\mathbb R^d}f(x)^2\,dx\cdot \int_{\mathbb R^d}g(x)^2\,dx.$$
Besides, there are Minkowski's inequalities for sums:
$$\left(\int_{\mathbb R^d}\left(f(x)+g(x)\right)^\frac1p\,dx\right)^p \le \left(\int_{\mathbb R^d}\left|f(x)\right|^\frac1p\,dx\right)^p+\left(\int_{\mathbb R^d}\left|g(x)\right|^\frac1p\,dx\right)^p.$$
As a special option for products, one can consider Laplace transform for convolution
$$\mathcal L\left\{\int_0^tf(t)g(x-t)dt\right\} = \mathcal L\left\{f(t)dt\right\}\cdot\mathcal L\left\{g(t)dt\right\}$$
