In this section, we develop a unified treatment of fractal Hilbert-type inequalities. In other words, we will establish a pair of general Hilbert-type inequalities that covers particular fractal inequalities presented in the introduction.
Our main result refers to a general kernel which is local fractional continuous on the fractal surface \((a,b)^{2}:=(a,b)\times (a,b)\).
Theorem 1
Let
\(\frac{1}{p}+\frac{1}{q}=1\), \(p>1\), \(0<\alpha \leq 1\), and let
\(K\in C_{\alpha }(a,b)^{2}\), \(\varphi ,\psi \in C_{\alpha }(a,b)\)
be non-negative functions. If the functions
F
and
G
are defined by
$$ F^{p} (x)= {}_{a}I_{b}^{\alpha } \bigl(K(x,y)\psi ^{-p}(y)\bigr), \qquad G^{q} (y)= {}_{a}I_{b}^{\alpha }\bigl(K(x,y)\phi ^{-q}(x)\bigr), $$
(9)
then, for all non-negative functions
\(f,g\in C_{\alpha }(a,b)\), the inequalities
$$ {}_{a}I_{b}^{\alpha } \bigl({}_{a}I_{b}^{\alpha }\bigl(K(x,y)f(x)g(y)\bigr) \bigr)\leq \bigl[ {}_{a}I_{b}^{\alpha }({\varphi }Ff)^{p} (x) \bigr]^{\frac{1}{p}} \bigl[{}_{a}I_{b}^{\alpha } \bigl( ({\psi }Gg)^{q} (y)\bigr) \bigr]^{\frac{1}{q}} $$
(10)
and
$$ {}_{a}I_{b}^{\alpha } \bigl( (G {\psi })^{-p}(y) \bigl[{}_{a}I_{b}^{ \alpha } \bigl(K(x,y)f(x)\bigr) \bigr]^{p} \bigr)\leq {{}_{a}I_{b}^{\alpha }} ( {\varphi }Ff)^{p} (x) $$
(11)
hold and are equivalent.
Proof
The left-hand side of inequality (10) can be rewritten as
$$\begin{aligned} &\frac{1}{\varGamma ^{2} (1+\alpha )} \int _{a}^{b} \int _{a}^{b} K(x,y)f(x)g(y) (dx)^{ \alpha }(dy)^{\alpha } \\ &\quad =\frac{1}{\varGamma ^{2} (1+\alpha )} \int _{a}^{b} \int _{a}^{b} K(x,y)f(x)\frac{ \varphi (x)}{\psi (y)}g(y) \frac{\psi (y)}{\varphi (x)}(dx)^{\alpha }(dy)^{ \alpha }, \end{aligned}$$
which is now suitable for the application of the Hölder inequality (7). Hence, we obtain
$$\begin{aligned} &\frac{1}{\varGamma ^{2} (1+\alpha )} \int _{a}^{b} \int _{a}^{b} K(x,y)f(x)g(y) (dx)^{ \alpha }(dy)^{\alpha } \\ &\quad \leq \biggl[\frac{1}{\varGamma ^{2} (1+\alpha )} \int _{a}^{b} \int _{a} ^{b} K(x,y)f^{p}(x) \frac{\varphi ^{p} (x)}{\psi ^{p} (y)}(dx)^{\alpha }(dy)^{ \alpha } \biggr]^{\frac{1}{p}} \\ &\qquad {}\times \biggl[\frac{1}{\varGamma ^{2} (1+\alpha )} \int _{a}^{b} \int _{a}^{b} K(x,y)g^{q}(y) \frac{\psi ^{q} (y)}{\varphi ^{q} (x)}(dx)^{ \alpha }(dy)^{\alpha } \biggr]^{\frac{1}{q}}. \end{aligned}$$
Now, by virtue of the Fubini theorem (see, e.g., [14]), we can switch the order of integration in the double integral, so by taking into account the definitions of functions F and G, we obtain (10), as claimed.
Our next step is to show the equivalence of inequalities (10) and (11). Therefore, suppose that inequality (10) holds and define the function g by
$$ g(y)= G^{-p}(y){\psi }^{-p}(y) \biggl[\frac{1}{\varGamma (1+\alpha )} \int _{a}^{b} K(x,y)f(x) (dx)^{\alpha } \biggr]^{p-1}. $$
Now, since \(\frac{1}{p}+\frac{1}{q}=1\), relation (10) implies the inequality
$$\begin{aligned}& \frac{1}{\varGamma (1+\alpha )} \int _{a}^{b} G^{-p}(y){\psi }^{-p}(y) \biggl[\frac{1}{\varGamma (1+\alpha )} \int _{a}^{b} K(x,y)f(x) (dx)^{ \alpha } \biggr]^{p} (dy)^{\alpha } \\& \quad = \frac{1}{\varGamma ^{2} (1+\alpha )} \int _{a}^{b} \int _{a}^{b} K(x,y)f(x)g(y) (dx)^{ \alpha }(dy)^{\alpha } \\& \quad \leq \biggl[\frac{1}{\varGamma (1+\alpha )} \int _{a}^{b} ({\varphi }Ff)^{p} (x) (dx)^{\alpha } \biggr]^{\frac{1}{p}} \biggl[ \frac{1}{ \varGamma (1+\alpha )} \int _{a}^{b} ({\psi }Gg)^{q} (y) (dy)^{\alpha } \biggr] ^{\frac{1}{q}} \\& \quad = \biggl[ \frac{1}{\varGamma (1+\alpha )} \int _{a}^{b} ({\varphi }Ff)^{p} (x) (dx)^{\alpha } \biggr]^{\frac{1}{p}} \\& \qquad {}\times \biggl[\frac{1}{\varGamma (1+\alpha )} \int _{a}^{b} (G \psi )^{q-pq}(y) \biggl[ \frac{1}{\varGamma (1+\alpha )} \int _{a}^{b} K(x,y)f(x) (dx)^{ \alpha } \biggr]^{q(p-1)} (dy)^{\alpha } \biggr]^{\frac{1}{q}}, \end{aligned}$$
which reduces to (11).
On the other hand, suppose that inequality (11) holds. Then yet another application of the Hölder inequality yields
$$\begin{aligned}& \frac{1}{\varGamma ^{2} (1+\alpha )} \int _{a}^{b} \int _{a}^{b} K(x,y)f(x)g(y) (dx)^{ \alpha }(dy)^{\alpha } \\& \quad = \frac{1}{\varGamma (1+\alpha )} \int _{a}^{b} \biggl[{\psi }^{-1}(y)G ^{-1}(y) \int _{a}^{b} K(x,y)f(x) (dx)^{\alpha } \biggr] \psi (y)G(y)g(y) (dy)^{ \alpha } \\& \quad \leq \biggl[\frac{1}{\varGamma (1+\alpha )} \int _{a}^{b} {\psi }^{-p}(y) G^{-p}(y) \biggl(\frac{1}{\varGamma (1+\alpha )} \int _{a}^{b} K(x,y)f(x) (dx)^{ \alpha } \biggr)^{p} (dy)^{\alpha } \biggr]^{\frac{1}{p}} \\& \qquad {} \times \biggl[\frac{1}{\varGamma (1+\alpha )} \int _{a}^{b} (\psi Gg)^{q}(y) (dy)^{\alpha } \biggr]^{\frac{1}{q}} \\& \quad \leq \biggl[\frac{1}{\varGamma (1+\alpha )} \int _{a}^{b} ({\varphi }Ff)^{p} (x) (dx)^{\alpha } \biggr]^{\frac{1}{p}} \biggl[ \frac{1}{ \varGamma (1+\alpha )} \int _{a}^{b} ({\psi }Gg)^{q} (y) (dy)^{\alpha } \biggr] ^{\frac{1}{q}}, \end{aligned}$$
which provides (10). Consequently, inequalities (10) and (11) are equivalent. □
It is not hard to see that our Theorem 1 covers fractal Hilbert-type inequalities (3) and (4) presented in the introduction. This follows by choosing a suitable power functions φ, ψ appearing in relations (10) and (11). However, this will not be done at this moment. Namely, considering kernels \(K_{1}(x,y)=1/\max \{x^{\alpha }, y^{\alpha }\}\) and appearing in (3) and (4), we see that they possess a common property, they are both homogeneous functions. Therefore, our next step is to derive consequence of Theorem 1 which refers to homogeneous kernels. The fractal Hilbert-type inequalities presented in the introduction will then follow as simple consequences of our next result.
Recall that the function \(K\in C_{\alpha }(0,\infty )^{2}\) is said to be homogeneous of degree \(-\alpha \lambda \), \(\lambda >0\), if \(K(tx,ty)=t^{-\alpha \lambda }K(x,y)\) for all \(t>0\). In order to formulate and prove the corresponding result, we need the following definition. For a non-negative function \(K\in C_{\alpha }(0,\infty )^{2}\), we define
$$ k_{\alpha }(\eta )= {}_{0}I_{\infty }^{\alpha } K(1,t)t^{-\alpha \eta }. $$
(12)
If nothing else is explicitly stated, we assume that the integral \(k_{\alpha }(\eta )\) converges for considered values of η. Now, we are ready to establish a pair of fractal Hilbert-type inequalities that correspond to a class of homogeneous kernels.
Theorem 2
Let
\(\frac{1}{p}+\frac{1}{q}=1\), \(p>1\), and let
\(f,g\in C_{\alpha }(0, \infty )\)
be non-negative functions. If
\(K\in C_{\alpha }(0,\infty )^{2}\)
is a non-negative homogeneous function of degree
\(-\alpha \lambda \), \(\lambda >0\), then the following inequalities hold:
$$\begin{aligned}& {{}_{0}I_{\infty }^{\alpha }} \bigl( {{}_{0}I_{\infty }^{\alpha }} \bigl(K(x,y)f(x)g(y)\bigr)\bigr) \\& \quad \leq L \bigl[ {{}_{0}I_{\infty }^{\alpha }} \bigl( x^{\alpha -\alpha \lambda +\alpha p(A_{1}-A_{2})} f^{p} (x) \bigr) \bigr]^{ \frac{1}{p}} \bigl[ {{}_{0}I_{\infty }^{\alpha }} \bigl( y^{\alpha - \alpha \lambda +\alpha q(A_{2}-A_{1})} g^{q}(y) \bigr) \bigr]^{ \frac{1}{q}} \end{aligned}$$
(13)
and
$$\begin{aligned}& {{}_{0}I_{\infty }^{\alpha }} \bigl[ y^{\alpha (p-1)(\lambda -1)+ \alpha p(A_{1}-A_{2})} \bigl({{}_{0}I_{\infty }^{\alpha }}\bigl(K(x,y)f(x)\bigr) \bigr) ^{p} \bigr] \\& \quad \leq L^{p} \bigl[ {{}_{0}I_{\infty }^{\alpha }} x^{\alpha -\alpha \lambda +\alpha p(A_{1}-A_{2})} f^{p} (x) \bigr], \end{aligned}$$
(14)
where
\(L=k_{\alpha }^{1/p}(pA_{2})k_{\alpha }^{1/q}(2-\lambda -qA_{1})\). In addition, relations (13) and (14) are equivalent.
Proof
We employ inequalities (10) and (11) with power functions \(\varphi (x)=x^{\alpha A_{1}}\) and \(\psi (y)=y^{\alpha A _{2}}\). Furthermore, making use of (9), it follows that
$$ F^{p} (x)=\frac{1}{\varGamma (1+\alpha )} \int _{0}^{\infty }K(x,y)y^{- \alpha pA_{2}}(dy)^{\alpha } $$
and
$$ G^{q} (x)=\frac{1}{\varGamma (1+\alpha )} \int _{0}^{\infty }K(x,y)x^{- \alpha qA_{1}}(dx)^{\alpha }. $$
In addition, since K is a homogeneous function of degree \(-\alpha \lambda \), \(\lambda >0\), a change of variables \(t=y/x\) provides
$$\begin{aligned} F^{p} (x)&=x^{\alpha -\alpha \lambda -\alpha pA_{2}} \frac{1}{\varGamma (1+\alpha )} \int _{0}^{\infty }K \biggl(1,\frac{y}{x} \biggr) \biggl(\frac{y}{x} \biggr) ^{-\alpha pA_{2}}\frac{1}{x^{\alpha }} (dy)^{\alpha } \\ &= x^{\alpha -\alpha \lambda -\alpha pA_{2}} \frac{1}{\varGamma (1+ \alpha )} \int _{0}^{\infty }K(1,t)t^{-\alpha pA_{2}}(dt)^{\alpha } \\ &=x^{\alpha -\alpha \lambda -\alpha pA_{2}}k_{\alpha }(pA_{2}) \end{aligned}$$
(15)
due to (6). Following the lines as in the previous step, we also obtain
$$ G^{q} (y)=y^{\alpha -\alpha \lambda -\alpha qA_{1}} \frac{1}{\varGamma (1+ \alpha )} \int _{0}^{\infty }K (t,1 )t^{-\alpha qA_{1}}(dt)^{ \alpha }. $$
Now, yet another application of the change of variables rule (6) with \(u=t^{-1}\), gives
$$\begin{aligned} G^{q} (y)&=-y^{\alpha -\alpha \lambda -\alpha qA_{1}}\frac{1}{\varGamma (1+\alpha )} \int _{0}^{\infty }K \biggl( 1,\frac{1}{t} \biggr) \biggl(\frac{1}{t} \biggr) ^{\alpha \lambda +\alpha qA_{1}-2\alpha } \bigl(-t^{-2\alpha }\bigr) (dt)^{ \alpha } \\ &=y^{\alpha -\alpha \lambda -\alpha qA_{1}}\frac{1}{\varGamma (1+\alpha )} \int _{0}^{\infty }K( 1,u) u^{\alpha \lambda +\alpha qA_{1}-2\alpha }(du)^{\alpha } \\ &= y^{\alpha -\alpha \lambda -\alpha qA_{1}}k_{\alpha }(2-\lambda -qA _{1}). \end{aligned}$$
(16)
Finally, inequalities (13) and (14) follow from relations (10), (11), (15), and (16). □
It should be noticed here that Theorem 2 holds for arbitrary parameters \(A_{1}\) and \(A_{2}\) such that the constant L and the integrals on the right-hand sides of (13) and (14) are convergent.
Generally speaking, we are not able to prove whether or not the constants L and \(L^{p}\) appearing on the right-hand sides of (13) and (14) are the best possible. However, it turns out that these constants are the best possible for a wide set of parameters \(A_{1}\), \(A_{2}\) and a weak condition on the kernel K. In order to establish the corresponding result, we first need the following lemma.
Lemma 1
Let
\(\lambda >0\), and let
\(\frac{1}{p}+\frac{1}{q}=1\), \(p>1\). If
\(K\in C_{\alpha }(0,\infty )^{2}\)
is a non-negative function such that
\(K(1,t)\)
is bounded on
\((0,1)\), then the following relation holds:
$$ {{}_{1}I_{\infty }^{\alpha }} \bigl[ x^{-\alpha (1+\varepsilon )} {{{}_{0}I_{1/x}^{\alpha }}} \bigl(t^{-\alpha pA_{2}-\frac{\varepsilon \alpha }{q}}K(1,t) \bigr) \bigr]\leq O(1),\quad \varepsilon \rightarrow 0^{+}, $$
(17)
where
\(A_{2}\leq \frac{1}{2p}\).
Proof
From the hypotheses, we have \(K(1,t)\leq C\) for some \(C>0\) and every \(t\in (0,1)\). Then it follows that
$$\begin{aligned} &{{}_{1}I_{\infty }^{\alpha }} \bigl[x^{-\alpha (1+\varepsilon )} {{{}_{0}I_{1/x}^{\alpha }}} \bigl(t^{-\alpha pA_{2}-\frac{\varepsilon \alpha }{q}}K(1,t) \bigr) \bigr] \\ &\quad \leq C {{}_{1}I_{\infty }^{\alpha }} \bigl[x^{-\alpha } {{{}_{0}I _{1/x}^{\alpha }} } \bigl(t^{-\frac{\alpha }{2}-\frac{\varepsilon \alpha }{q}} \bigr) \bigr]. \end{aligned}$$
(18)
Furthermore, utilizing the change of variables rule (6) with \(g(t)=t^{\frac{1}{2}-\frac{\varepsilon }{q}}\), \([g'(t)]^{\alpha }= (\frac{1}{2}-\frac{\varepsilon }{q} )^{\alpha }t^{-\frac{ \alpha }{2}-\frac{\varepsilon \alpha }{q}}\), we obtain
$$\begin{aligned} &{{}_{1}I_{\infty }^{\alpha }} \bigl[x^{-\alpha } \bigl({{{}_{0}I_{1/x} ^{\alpha }}} \bigl(t^{-\frac{\alpha }{2}- \frac{\varepsilon \alpha }{q}} \bigr) \bigr) \bigr] \\ &\quad =\frac{1}{\varGamma ^{2} (1+\alpha )} \int _{1}^{\infty }x^{-\alpha } \biggl( \int _{0}^{1/x}t^{-\frac{\alpha }{2}- \frac{\varepsilon \alpha }{q}} (dt)^{\alpha } \biggr) (dx)^{\alpha } \\ &\quad =\frac{1}{\varGamma ^{2} (1+\alpha )} \int _{1}^{\infty }x^{-\alpha } \biggl( \frac{1}{ (\frac{1}{2}-\frac{\varepsilon }{q} ) ^{\alpha }} \int _{0}^{x^{-\frac{1}{2}+\frac{\varepsilon }{q}}}(du)^{ \alpha } \biggr) (dx)^{\alpha } \\ &\quad =\frac{1}{\varGamma ^{2} (1+\alpha ) (\frac{1}{2}-\frac{\varepsilon }{q} )^{\alpha }} \int _{1}^{\infty }x^{-\frac{3\alpha }{2}+\frac{ \varepsilon \alpha }{q}}(dx)^{\alpha }. \end{aligned}$$
Now, if \(g(x)=x^{-\frac{1}{2}+\frac{\varepsilon }{q}}\), then \([g'(x)]^{\alpha }=- (\frac{1}{2}-\frac{\varepsilon }{q} ) ^{-\alpha } x^{-\frac{3\alpha }{2}+\frac{\varepsilon \alpha }{q}}\), so by (6), we have
$$ {{}_{1}I_{\infty }^{\alpha }} \bigl[x^{-\alpha } {{{}_{0}I_{1/x}^{\alpha }} } \bigl(t^{-\frac{\alpha }{2}-\frac{\varepsilon \alpha }{q}} \bigr) \bigr] =\frac{1}{ \varGamma ^{2} (1+\alpha ) (\frac{1}{2}-\frac{\varepsilon }{q} ) ^{2\alpha }}. $$
(19)
Finally, combining (18) and (19), we obtain (17), as claimed. □
Our next intention is to impose the condition on parameters \(A_{1}\) and \(A_{2}\) for which the constants appearing on the right-hand sides of inequalities (13) and (14) are the best possible. It should be noticed here that if
$$ pA_{2}+qA_{1}=2-\lambda , $$
(20)
then the constant L from Theorem 2 reduces to the form without exponents, i.e.,
$$ L^{*}=k_{\alpha }(pA_{2}). $$
(21)
We will show that if the parameters \(A_{1}\) and \(A_{2}\) are related by (20), then the constants appearing on the right-hand sides of (13) and (14) are the best possible. In fact, if (20) holds, inequalities (13) and (14) reduce to
$$\begin{aligned} &{{}_{0}I_{\infty }^{\alpha }} \bigl( {{}_{0}I_{\infty }^{\alpha }} \bigl(K(x,y)f(x)g(y)\bigr)\bigr) \\ &\quad \leq L^{*} \bigl[ {{}_{0}I_{\infty }^{\alpha }} \bigl( x^{\alpha pqA _{1}-\alpha } f^{p} (x) \bigr) \bigr]^{\frac{1}{p}} \bigl[ {{}_{0}I _{\infty }^{\alpha }} \bigl( y^{\alpha pqA_{2}-\alpha } g^{q}(y) \bigr) \bigr]^{\frac{1}{q}} \end{aligned}$$
(22)
and
$$\begin{aligned} &{{}_{0}I_{\infty }^{\alpha }} \bigl[ y^{\alpha p(\lambda -1)+\alpha pqA_{1}} \bigl({{}_{0}I_{\infty }^{\alpha }} \bigl(K(x,y)f(x)\bigr) \bigr)^{p} \bigr] \\ &\quad \leq \bigl(L^{*}\bigr)^{p} \bigl[ {{}_{0}I_{\infty }^{\alpha }} \bigl(x^{\alpha pqA _{1}-\alpha } f^{p} (x)\bigr) \bigr], \end{aligned}$$
(23)
where \(L^{*}\) is defined by (21).
Theorem 3
Suppose that
\(\frac{1}{p}+\frac{1}{q}=1\), \(p>1\), and let
\(f,g\in C _{\alpha }(0,\infty )\)
be non-negative functions. Further, let
\(K\in C_{\alpha }(0,\infty )^{2}\)
be a non-negative homogeneous function of degree
\(-\alpha \lambda \), \(\lambda >0\), such that
\(K(1,t)\)
is bounded on
\((0,1)\). If the parameters
\(A_{1}\)
and
\(A_{2}\)
satisfy relation
\(pA_{2}+qA_{1}=2-\lambda \), then the constants
\(L^{*}\)
and
\((L^{*})^{p}\)
appearing on the right-hand sides of (22) and (23) are the best possible.
Proof
Let \(f(x)=x^{-\alpha qA_{1}-\frac{\varepsilon \alpha }{p}} \chi _{[1,\infty )}(x)\) and \(g(y)=y^{-\alpha pA_{2}-\frac{\varepsilon \alpha }{q}}\chi _{[1,\infty )}(y)\), where \(\chi _{A}\) stands for a characteristic function of a set A. Now, let us suppose that there exists a smaller constant \(0< M< L^{\ast }\) such that inequality (22) holds. Denote by J the right-hand side of inequality (22). Then, with the above defined functions f and g, we have
$$\begin{aligned} J&=M \biggl(\frac{1}{\varGamma (1+\alpha )} \int _{1}^{\infty }x^{-\alpha \varepsilon -\alpha }(dx)^{\alpha } \biggr)^{\frac{1}{p}} \biggl(\frac{1}{ \varGamma (1+\alpha )} \int _{1}^{\infty }y^{-\alpha \varepsilon -\alpha }(dy)^{ \alpha } \biggr)^{\frac{1}{q}} \\ &=\frac{M}{\varepsilon ^{\alpha }\varGamma (1+\alpha )}. \end{aligned}$$
(24)
Further, utilizing substitution \(t=\frac{y}{x}\) and taking into account Lemma 1, we obtain the following estimate:
$$\begin{aligned} &{{}_{0}I_{\infty }^{\alpha }} \bigl( {{}_{0}I_{\infty }^{\alpha }} \bigl(K(x,y)f(x)g(y)\bigr)\bigr) \\ &\quad = {{}_{1}I_{\infty }^{\alpha }} \bigl[ x^{-\alpha qA_{1}-\frac{ \alpha \varepsilon }{p}} {}_{1}I_{\infty }^{\alpha } \bigl(y^{-\alpha pA_{2}-\frac{\alpha \varepsilon }{q}}K(x,y)\bigr) \bigr] \\ &\quad = {{}_{1}I_{\infty }^{\alpha }} \bigl[ x^{-\alpha (1+\varepsilon )} \bigl( {{}_{0}I_{\infty }^{\alpha }} \bigl(t^{-\alpha pA_{2}-\frac{\alpha \varepsilon }{q}}K(1,t)\bigr) - {} _{0}I_{1/x}^{\alpha } \bigl(t^{-\alpha pA_{2}-\frac{\alpha \varepsilon }{q}}K(1,t)\bigr) \bigr) \bigr] \\ &\quad \geq \frac{1}{\varepsilon ^{\alpha }\varGamma (1+\alpha )} \biggl( k _{\alpha } \biggl(pA_{2}+ \frac{\varepsilon }{q} \biggr)+o(1) \biggr). \end{aligned}$$
(25)
Moreover, from (22), (24), and (25), we get
$$ k_{\alpha } \biggl(pA_{2}+\frac{\varepsilon }{q} \biggr)+o(1)\leq M. $$
(26)
Now, by letting \(\varepsilon \rightarrow 0^{+}\), it follows that relation (26) contradicts with our assumption \(M< L^{*}=k_{ \alpha }(pA_{2})\).
Finally, equivalence of inequalities (22) and (23) means that the constant \((L^{\ast })^{p} = [k_{\alpha }(pA_{2}) ] ^{p}\) is also the best possible in (23). The proof is now completed. □