# An inequality by using general Hölder's inequality

Let $w$ be a weight function and $\frac{1}{p}=\frac{1}{p_1}+\frac{1}{p_2}+\dots+\frac{1}{p_n}$. For suitable functions let $\overrightarrow{f}=(f_1,f_2,\dots,f_n)$ and operator $T$ satisifies the condition $$T(\overrightarrow{f})(x)\leq \prod_{i=1}^{n}f_{i}(x).\tag{*}$$ By using (*) and generalization of Hölder's inequality, I get $$\left(\int_{\mathbb{R}^n}|T(\overrightarrow{f})(x)w(x)|^pdx\right)^{\frac{1}{p}}\leq \left(\int_{\mathbb{R}^n}| \prod_{i=1}^{n}f_{i}(x)w(x)|^pdx\right)^{\frac{1}{p}}\leq \prod_{i=1}^{n}\left(\int_{\mathbb{R}^n}|f_{i}(x)|^{p_i}w(x)^pdx\right)^{\frac{1}{p_i}},$$ where I apply Hölder's inequality for the measure $d\mu(x)=w(x)^p$.

But the paper which I read now writes this inequality as following $$\left(\int_{\mathbb{R}^n}|T(\overrightarrow{f})(x)w(x)|^pdx\right)^{\frac{1}{p}}\leq \left(\int_{\mathbb{R}^n}| \prod_{i=1}^{n}f_{i}(x)w(x)|^pdx\right)^{\frac{1}{p}}\leq \prod_{i=1}^{n}\left(\int_{\mathbb{R}^n}|f_{i}(x)w(x)|^{p_i}dx\right)^{\frac{1}{p_i}}.$$ Which one is true?

• Yours is definitely correct, I cannot obtain the second. If one modifies the second inequality such that $|w|^{p_i}$ is included in exactly one of the $n$ integrands then one can get something similar to the second. (i.e., applying Hoelder to the product of $|f_1w|^p, |f_2|^p \dots |f_n|^p$). – daw Nov 24 '17 at 12:55
• I am pretty sure that what you want to prove is that: $$\|Tf\|_{L^p(w)} \le \prod_{i=1}^{n}\left(\int_{\mathbb{R}^n}|f_{i}(x)|^{p_i}w(x)^pdx\right)^{\frac{1}{p_i}},$$ Am I correct? – Guy Fsone Nov 24 '17 at 14:18
• @GuyFsone I wonder only which one of the inequalities true? Am i right or wrong? – user315531 Nov 24 '17 at 14:26
• @user315531tHEN your result is correct. I have added and additioanl remark below my answer. – Guy Fsone Nov 24 '17 at 14:40

Your result is absolutely true:

Here is a way to check it: Let $\mu$ be the measure $$d\mu(x)= w^p(x)dx$$

Now we apply the Holder inequality w.r.t the measure $\mu$ then we have,

$$\left(\int_{\mathbb{R}^n}|T(\overrightarrow{f})(x)|^p\color{red}{d\mu(x)}\right)^{\frac{1}{p}}\leq \left(\int_{\mathbb{R}^n}| \prod_{i=1}^{n}f_{i}(x)|^p\color{red}{d\mu(x)}\right)^{\frac{1}{p}}\leq \prod_{i=1}^{n}\left(\int_{\mathbb{R}^n}|f_{i}(x)|^{p_i}\color{red}{d\mu(x)}\right)^{\frac{1}{p_i}},$$

replacing $\mu$ by its value we get: $$\left(\int_{\mathbb{R}^n}|T(\overrightarrow{f})(x)w(x)|^pdx\right)^{\frac{1}{p}}\leq \left(\int_{\mathbb{R}^n}\color{red}{\left|\prod_{i=1}^{n}f_{i}(x)\right|^p} w^p(x)dx\right)^{\frac{1}{p}}\leq \prod_{i=1}^{n}\left(\int_{\mathbb{R}^n}|f_{i}(x)|^{p_i}w(x)^pdx\right)^{\frac{1}{p_i}},$$

Which provide your result.

Warning: However, as $\color{blue}{xa_1\times xa_2\times xa_3 = x^3\times a_1\times a_2\times a_3}$, note that there might be confusion between the following: $$\left|\prod_{i=1}^{n}f_{i}(x)w(x)\right|^p = w^{\color{blue}{np}}(x)\left|\prod_{i=1}^{n}f_{i}(x)\right|^p$$ and $$w^p(x)\left|\prod_{i=1}^{n}f_{i}(x)\right|^p =\left|\prod_{i=1}^{n}f_{i}(x)\right|^p\times w^p(x)$$

• coloring the things like you do is the hell of a strategy to teach, nice!!!! – user499752 Nov 25 '17 at 2:47
• @AlfredfromBatman the colors make the contain more accessible I think – Guy Fsone Nov 27 '17 at 13:20

Use that: $\left(\int_{\mathbb{R}^n}| \prod_{i=1}^{n}f_{i}(x)w(x)|^pdx\right)^{\frac{1}{p}}\leq \prod_{i=1}^{n}\left(\int_{\mathbb{R}^n}|f_{i}(x)|^{p_i}w(x)^pdx\right)^{\frac{1}{p_i}}$

We can rename $\tilde{f}_i(x) = f_i(x)w(x)$ and $\tilde{w}(x) \equiv 1$ and get

$\left(\int_{\mathbb{R}^n}| \prod_{i=1}^{n}f_{i}(x)w(x)|^pdx\right)^{\frac{1}{p}} \\ = \left(\int_{\mathbb{R}^n}| \prod_{i=1}^{n}\tilde{f}_{i}(x)\tilde{w}(x)|^pdx\right)^{\frac{1}{p}} \\ \leq \prod_{i=1}^{n}\left(\int_{\mathbb{R}^n}|\tilde{f}_{i}(x)|^{p_i}\tilde{w}(x)^pdx\right)^{\frac{1}{p_i}} \\ = \prod_{i=1}^{n}\left(\int_{\mathbb{R}^n}|f_{i}(x)w(x)|^{p_i}dx\right)^{\frac{1}{p_i}}$

Edit: So I'd say both inequalities are true.

• I am not sure. When i write explictly I found $\left(\int_{\mathbb{R}^n}| f_{1}(x)f_{2}(x)\dots f_{n}(x)|^p w(x)^p dx\right)^{\frac{1}{p}}$ but i think you find $\left(\int_{\mathbb{R}^n}| f_{1}(x)w(x)f_{2}(x)w(x)\dots f_{n}(x)w(x)|^pdx\right)^{\frac{1}{p}}$ and i am not sure this one is true? – user315531 Nov 24 '17 at 14:40
• I think this $\tilde{w}(x) \equiv 1$ is faking. it does not have to be there. just remove it . It is pointless. – Guy Fsone Nov 24 '17 at 14:51
• Also the first equality in your estimate is no true. see this remark: $\color{blue}{xa_1\times xa_2\times xa_3 = x^3\times a_1\times a_2\times a_3}$ I have appointed this in my answer. – Guy Fsone Nov 24 '17 at 14:54