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! :)

  • $\begingroup$ You can check out Ladyzhenskaya's and Nash's inequalities, although they deal only with one function and its gradient. $\endgroup$ – Svetoslav Sep 8 '15 at 16:08
  • $\begingroup$ There is a more general form, involving complementary Young's functions... and check out books with "Inequalities" in titles: "Inequalities" by Beckenbach-Bellman, "Inequalities" by Hardy-Littlewood-Polya, and "Cauchy-Schwarz master class" by Steele. $\endgroup$ – user147263 Sep 10 '15 at 6:11
  • $\begingroup$ See the following post: Generalized Hölder inequality. $\endgroup$ – user 170039 Oct 31 '17 at 13:48

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 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 $$\mathcal L\left\{\int_0^tf(t)g(x-t)dt\right\} = \mathcal L\left\{f(t)dt\right\}\cdot\mathcal L\left\{g(t)dt\right\}$$

  • $\begingroup$ I need it for products. A better inequality than Hölder if there is one. $\endgroup$ – Martingalo Feb 7 '17 at 23:56
  • $\begingroup$ Equality can satisfy to non-strict inequality $\endgroup$ – Yuri Negometyanov Feb 8 '17 at 0:40

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