An extended differential form of Hilbert’s inequality

Lemma 2.1 see Azar et al., 2014, Lemma 2.2

Let $r>1\text{,}\frac{1}{r}+\frac{1}{s}=1\text{,}\phi (x)>0$ , the derivatives ${\phi }^{\prime }\text{,}{\phi }^{″}\text{,}\cdots \text{,}{\phi }^{(n)}$ exists and positive and ${\phi }^{(n)}\in L(0\text{,}\infty )\text{,}(n=0\text{,}1\text{,}2\text{,}\cdots )\text{,}({\phi }^{(0)}≔\phi )\text{,}$ where $L^p$ denotes the space of all Lebesgue integrable functions (If $p=1$ we obtain L), Moreover, suppose that $\phi (0)={\phi }^{″}(0)={\phi }^{″}(0)=\cdots ={\phi }^{(n-1)}(0)=0$ , then for $t\text{,}\alpha >0$ we have

Let $0<r<1\text{,}\frac{1}{r}+\frac{1}{s}=1$ , Let $\phi$ be as in Lemma 2.2, then for $t>0$ and $\beta \in \mathbb{R}$ $(\beta s+1>0)$ , we get

Using integration by parts n times, we get $${\int }_0^{\infty }{e}^{-\mathit{tx}}\phi (x)\mathit{dx}=\frac{1}{t^n}{\int }_0^{\infty }{e}^{-\mathit{tx}}{\phi }^{(n)}(x))\mathit{dx}\text{.}$$

Applying reverse Hölder inequality, then use (2.1), we obtain $$\begin{array}{cc}\hfill {\int }_0^{\infty }{e}^{-\mathit{tx}}\phi (x)\mathit{dx}=& \frac{1}{t^n}{\int }_0^{\infty }(x^{\beta }{e}^{-\frac{\mathit{tx}}{s}})(x^{-\beta }{e}^{-\frac{\mathit{tx}}{r}}{\phi }^{(n)}(x))\mathit{dx}\text{.}\hfill \\ \hfill ⩾& \frac{1}{t^n}{\left({\int }_0^{\infty }{x}^{\beta s}{e}^{-\mathit{xt}}\mathit{dx}\right)}^{\frac{1}{s}}{\left({\int }_0^{\infty }{x}^{-\beta r}{e}^{-\mathit{xt}}{(({\phi }^{(n)})(x))}^r\mathit{dx}\right)}^{\frac{1}{r}}\hfill \\ \hfill =& \frac{1}{t^n}{\left(\frac{1}{t^{\beta s+1}}\mathrm{\Gamma }(\beta s+1)\right)}^{\frac{1}{s}}{\left({\int }_0^{\infty }{x}^{-\beta r}{e}^{-\mathit{xt}}{(({\phi }^{(n)})(x))}^r\mathit{dx}\right)}^{\frac{1}{r}}\hfill \end{array}$$ this leads to (2.5). □

Let $p>1\text{,}\frac{1}{p}+\frac{1}{q}=1\text{,}f(x)\text{,}g(y)>0\text{,}\lambda >2n\text{,}\gamma \in (n-\frac{\lambda }{p}\text{,}\frac{\lambda }{q}-n)\text{,}n=0\text{,}1\text{,}\dots$ and $f(x)\text{,}g(y)$ satisfies the conditions of Lemma 2.1 such that

${\int }_0^{\infty }{x}^{p(n+1)-p\gamma -\lambda -1}{(f^{(n)}(x))}^p\mathit{dx}<\infty$ , ${\int }_0^{\infty }{y}^{q(n+1)+q\gamma -\lambda -1}{(g^{(n)}(y))}^q\mathit{dy}<\infty$ , then:

If we use the relation (2.3) and apply the Hölder’s inequality, we obtain

Using Lemma 2.2 for $r=p\text{,}s=q\text{,}\alpha ={\alpha }_1$ , and then for $r=q\text{,}s=p\text{,}\alpha ={\alpha }_2$ we obtain respectively, $${\left({\int }_0^{\infty }{e}^{-\mathit{tx}}f(x)\mathit{dx}\right)}^p⩽t^{-\mathit{np}-{\alpha }_{1}p-p+1}\mathrm{\Gamma }{({\alpha }_{1}q+1)}^{\frac{p}{q}}{\int }_0^{\infty }{x}^{-{\alpha }_{1}p}{e}^{-\mathit{tx}}{(f^{(n)}(x))}^p\mathit{dx}$$ and $${\left({\int }_0^{\infty }{e}^{-\mathit{ty}}g(y)\mathit{dy}\right)}^q⩽t^{-\mathit{nq}-{\alpha }_{2}q-q+1}\mathrm{\Gamma }{({\alpha }_{2}p+1)}^{\frac{q}{p}}{\int }_0^{\infty }{y}^{-{\alpha }_{2}q}{e}^{-\mathit{ty}}{(g^{(n)}(y))}^q\mathit{dy}$$

If we substitute these two inequalities in (3.2) we get $$I⩽\frac{\mathrm{\Gamma }{({\alpha }_{1}q+1)}^{\frac{1}{q}}\mathrm{\Gamma }{({\alpha }_{2}p+1)}^{\frac{1}{p}}}{\mathrm{\Gamma }(\lambda )}{\left({\int }_0^{\infty }{x}^{-{\alpha }_{1}p}{(f^{(n)}(x))}^p\left({\int }_0^{\infty }{t}^{\lambda +p(\gamma -n-{\alpha }_1-1)}{e}^{-\mathit{tx}}\mathit{dt}\right)\mathit{dx}\right)}^{\frac{1}{p}}\times {\left({\int }_0^{\infty }{t}^{-{\alpha }_{2}q}{(g^{(n)}(y))}^q\left({\int }_0^{\infty }{t}^{\lambda -q(\gamma +n+{\alpha }_2+1)}{e}^{-\mathit{ty}}\mathit{dt}\right)\mathit{dy}\right)}^{\frac{1}{q}}=C_1{\left({\int }_0^{\infty }{x}^{p(n+1)-p\gamma -\lambda -1}{(f^{(n)}(x))}^p\mathit{dx}\right)}^{\frac{1}{p}}{\left({\int }_0^{\infty }{y}^{q(n+1)+q\gamma -\lambda -1}{(g^{(n)}(y))}^q\mathit{dy}\right)}^{\frac{1}{q}}\text{,}$$ where $C_1=\frac{\mathrm{\Gamma }{({\alpha }_{1}q+1)}^{\frac{1}{q}}\mathrm{\Gamma }{({\alpha }_{2}p+1)}^{\frac{1}{p}}\mathrm{\Gamma }{(\lambda +p(\gamma -n-{\alpha }_1-1)+1)}^{\frac{1}{p}}\mathrm{\Gamma }{(\lambda -q(\gamma +n+{\alpha }_2+1)+1)}^{\frac{1}{q}}}{\mathrm{\Gamma }(\lambda )}$ . Now, let ${\alpha }_1=\frac{\lambda +p(\gamma -n-1)}{\mathit{pq}}$ and ${\alpha }_2=\frac{\lambda -q(\gamma +n+1)}{\mathit{pq}}$ , we obtain

$C_1=C=\frac{\mathrm{\Gamma }(\frac{\lambda }{p}+\gamma -n)\mathrm{\Gamma }(\frac{\lambda }{q}-\gamma -n)}{\mathrm{\Gamma }(\lambda )}$ . Hence, Inequality (3.1) is proved. It remains to show that the constant C is the best possible. To do that we define two functions: $$f_{\epsilon }(x)=\left\{\begin{array}{cc}0\text{,}\hfill & 0<x<1\hfill \\ \frac{\mathrm{\Gamma }\left(\frac{\lambda +q\gamma -\epsilon }{p}-n\right)}{\mathrm{\Gamma }\left(\frac{\lambda +p\gamma -\epsilon }{p}\right)}{x}^{\frac{\lambda +p\gamma -\epsilon }{p}-1\text{,}}\hfill & x⩾1\hfill \end{array}\right.$$ and $$g_{\epsilon }(y)=\left\{\begin{array}{cc}0\text{,}\hfill & 0<y<1\hfill \\ \frac{\mathrm{\Gamma }(\frac{\lambda -q\gamma -\epsilon }{q}-n)}{\mathrm{\Gamma }(\frac{\lambda -q\gamma -\epsilon }{q})}{y}^{\frac{\lambda -q\gamma -\epsilon }{q}-1}\text{,}\hfill & y⩾1\hfill \end{array}\text{,}\right.$$ where $0<\epsilon <\min \{\lambda -q(\gamma +n)\text{,}\lambda +p(\gamma -n)\}$ . Thus, we find $f_{\epsilon }^{(n)}(x)=x^{\frac{\lambda +p\gamma -\epsilon }{p}-n-1}$ , $x>1$ , and $g_{\epsilon }^{(n)}(y)=y^{\frac{\lambda -q\gamma -\epsilon }{q}-n-1}$ , $y>1$ .

Suppose that $C=\frac{\mathrm{\Gamma }\left(\frac{\lambda }{p}+\gamma -n\right)\mathrm{\Gamma }\left(\frac{\lambda }{q}-\gamma -n\right)}{\mathrm{\Gamma }(\lambda )}$ is not the best possible, then there exist $0<k<C$ such that

On the other hand, we obtain (the constant $D=\frac{\mathrm{\Gamma }\left(\frac{\lambda +p\gamma -\epsilon }{p}-n\right)\mathrm{\Gamma }\left(\frac{\lambda -q\gamma -\epsilon }{q}-n\right)}{\mathrm{\Gamma }\left(\frac{\lambda +p\gamma -\epsilon }{p}\right)\mathrm{\Gamma }\left(\frac{\lambda -q\gamma -\epsilon }{q}\right)}$ )

If we consider (3.3), (3.4) and the relation between the Beta and Gamma functions $B(u\text{,}v)=\frac{\mathrm{\Gamma }(u)\mathrm{\Gamma }(v)}{\mathrm{\Gamma }(u+v)}$ and let $\epsilon \to 0^{+}$ , we get $$k⩾\frac{\mathrm{\Gamma }\left(\frac{\lambda +p\gamma }{p}-n\right)\mathrm{\Gamma }(\frac{\lambda -q\gamma }{q}-n)}{\mathrm{\Gamma }\left(\frac{\lambda }{p}+\gamma \right)\mathrm{\Gamma }\left(\frac{\lambda }{q}-\gamma \right)}\beta \left(\frac{\lambda }{q}-\gamma \text{,}\frac{\lambda }{p}+\gamma \right)=\frac{\mathrm{\Gamma }\left(\frac{\lambda +p\gamma }{p}-n\right)\mathrm{\Gamma }\left(\frac{\lambda -q\gamma }{q}-n\right)}{\mathrm{\Gamma }(\lambda )}\text{,}$$ which is in contradiction with our assumption. The theorem is proved. □

If $f\text{,}g>0$ , $0<p<1\text{,}\frac{1}{p}+\frac{1}{q}=1\text{,}\lambda >2n\text{,}\gamma \in \left(n-\frac{\lambda }{p}\text{,}\frac{\lambda }{q}-n\right)\text{,}n=0\text{,}1\text{,}\dots$ and $f\text{,}g$ satisfies the conditions of Lemma 2.1 such that ${\int }_0^{\infty }{x}^{p(n+1)-p\gamma -\lambda -1}{(f^{(n)}(x))}^p\mathit{dx}<\infty$ , and ${\int }_0^{\infty }{y}^{q(n+1)+q\gamma -\lambda -1}{(g^{(n)}(y))}^q\mathit{dy}<\infty$ , then we have the reverse form of (3.1) as

Using (2.3) and the reverse Hölder inequality, we have

If we use Lemma 2.2 for $r=p\text{,}s=q$ , $\beta ={\beta }_1$ and then for $r=q\text{,}s=p\text{,}\beta ={\beta }_2$ respectively, we find $${\left({\int }_0^{\infty }{e}^{-\mathit{tx}}f(x)\mathit{dx}\right)}^p⩾t^{-\mathit{np}-\frac{p}{q}-{\beta }_{1}p}\mathrm{\Gamma }{({\beta }_{1}q+1)}^{\frac{p}{q}}{\int }_0^{\infty }{x}^{-{\beta }_{1}p}{e}^{-\mathit{tx}}{(f^{(n)}(x))}^p\mathit{dx}\text{,}$$ and $${\left({\int }_0^{\infty }{e}^{-\mathit{ty}}f(y)\mathit{dy}\right)}^q⩾t^{-\mathit{nq}-\frac{q}{p}-{\beta }_{2}q}\mathrm{\Gamma }{({\beta }_{1}q+1)}^{\frac{q}{p}}{\int }_0^{\infty }{y}^{-{\beta }_{2}q}{e}^{-\mathit{ty}}{(f^{(n)}(y))}^q\mathit{dy}\text{.}$$

If we substitute the last two inequalities in (3.6) and make some computations and then letting: ${\beta }_1=\frac{\lambda +p(\gamma -n-1)}{\mathit{pq}}$ and ${\beta }_2=\frac{\lambda -q(\gamma +n+1)}{\mathit{pq}}$ we arrive at inequality (3.5). To prove that the constant is the best possible, we define $f_{\epsilon }(x)$ and $g_{\epsilon }(y)$ as in the proof of Theorem 3.1. If C is not the best possible one, then there is a positive number $k⩾C$ such that (3.6) is still valid as we replace C by $k$ , then

On the other hand, we find

If we let $\epsilon \to 0^{+}$ , from (3.7) to (3.8) we find $k⩾C$ . the theorem is proved. □

If we put $n=0$ in (3.1) we get

Note that if we set in (3.9) $\gamma =\frac{p-\lambda -{\mathit{pqA}}_1}{p}\left(-\frac{\lambda }{p}<\gamma <\frac{\lambda }{q}\right)$ under the condition ${\mathit{pA}}_2+{\mathit{qA}}_1=2-\lambda$ we obtain inequality (1.2) from the introduction. Similarly, we may obtain the reverse form of (1.2) from (3.5).