A multiple Hilbert-type integral inequality with a non-homogeneous kernel

If $p>1$, $\frac{1}{p}+\frac{1}{q}=1$, $f(\ge 0)\in L^p(0,\mathrm{\infty })$, $g(\ge 0)\in L^q(0,\mathrm{\infty })$, ${\parallel f\parallel }_p,{\parallel g\parallel }_q>0$, then we have the following equivalent inequalities (cf. [1]):

(1)

(2)

where the constant factor $\frac{\pi }{sin(\pi /p)}$ is the best possible. (1) is the well-known Hardy-Hilbert integral inequality. Define the Hardy-Hilbert integral operator $T:L^p(0,\mathrm{\infty })\to L^p(0,\mathrm{\infty })$ as follows: for $f\in L^p(0,\mathrm{\infty })$, $Tf(y):={\int }_0^{\mathrm{\infty }}\frac{1}{x+y}f(x)dx$ ($y\in (0,\mathrm{\infty })$).

Then in view of (2), it follows ${\parallel Tf\parallel }_p<\frac{\pi }{sin(\pi /p)}{\parallel f\parallel }_p$ and $\parallel T\parallel \le \frac{\pi }{sin(\pi /p)}$. Since the constant is the best possible, we find $\parallel T\parallel =\frac{\pi }{sin(\pi /p)}$.

where the constant factor $k_{\lambda }$ is the best possible. For $n=2$, $k_{\lambda }(x,y)=\frac{1}{x^{\lambda }+y^{\lambda }}$ in (5), we obtain (3). Inequality (5) is an extension of the results in [9–12] and [13]. In recent years, [14] and [15] considered some Hilbert-type operators relating (1)-(3); [16] also considered a multiple Hilbert-type integral operator with the homogeneous kernel of $-n+1$-degree and the relating particular case of (5) (for $\lambda =n-1$, $\frac{1}{r_i}=\frac{1}{n-1}(1-\frac{1}{p_i})$).

In this paper, by using the way of weight functions and the technic of real analysis, a multiple Hilbert-type integral inequality with a non-homogeneous kernel is given. The operator expression with the norm, the reverses and some examples with the particular kernels are considered.

Proof

and then (6) is valid. □

Definition 1 If $n\in \mathbf{N}$, ${\mathbf{R}}_{+}^n:=\{(x_1,\dots ,x_n)|x_i>0(i=1,\dots ,n)\}$, $\lambda \in \mathbf{R}$, $k_{\lambda }(x_1,\dots ,x_n)$ is a measurable function in ${\mathbf{R}}_{+}^n$ such that for any $u>0$ and $(x_1,\dots ,x_n)\in {\mathbf{R}}_{+}^n$, $k_{\lambda }(u{x}_1,\dots ,u{x}_n)=u^{-\lambda }{k}_{\lambda }(x_1,\dots ,x_n)$, then we call $k_{\lambda }(x_1,\dots ,x_n)$ the homogeneous function of −λ-degree in ${\mathbf{R}}_{+}^n$.

Setting $u_j=x_j/x_i$ ($j\ne i,n$) and $u_n=x_n^{\mathrm{\prime }}/x_i$ in the above integral, we find ${\omega }_i(x_i)=H(i)=H(n)=k_{\lambda }$. □

is finite in a neighborhood of $({\lambda }_1,\dots ,{\lambda }_{n-1})$ if any only if $k({\tilde{\lambda }}_1,\dots ,{\tilde{\lambda }}_{n-1})$ is continuous at $({\lambda }_1,\dots ,{\lambda }_{n-1})$.

Proof The sufficiency property is obvious. We prove the necessary property of the condition by mathematical induction in the following. For $n=2$, there exists $I:=\{{\tilde{\lambda }}_1|{\tilde{\lambda }}_1={\lambda }_1+{\delta }_1,|{\delta }_1|\le {\delta }_0,{\delta }_0>0\}$ such that for any ${\tilde{\lambda }}_1\in I$, $k({\tilde{\lambda }}_1)\in \mathbf{R}$. Since for ${\tilde{\lambda }}_1={\lambda }_1+{\delta }_1\in I$ (${\delta }_1\ne 0$),

and $k({\lambda }_1-{\delta }_0)+k({\lambda }_1+{\delta }_0)<\mathrm{\infty }$, then by the Lebesgue control convergence theorem (cf. [17]), it follows $k({\lambda }_1+{\delta }_1)=k({\lambda }_1)+o(1)$ (${\delta }_1\to 0$). Assuming that for n (≥2), $k({\tilde{\lambda }}_1,\dots ,{\tilde{\lambda }}_{n-1})$ is continuous at $({\lambda }_1,\dots ,{\lambda }_{n-1})$, then for $n+1$, by the result of $n=2$, since $k({\lambda }_1+{\delta }_1,\dots ,{\lambda }_n+{\delta }_n)$ is finite in a neighborhood of $({\lambda }_1,\dots ,{\lambda }_n)$, we find

By mathematical induction, we prove that for $n\in \mathbf{N}\setminus \{1\}$, $k({\tilde{\lambda }}_1,\dots ,{\tilde{\lambda }}_{n-1})$ is continuous at $({\lambda }_1,\dots ,{\lambda }_{n-1})$. □

Without loss of generality, we estimate the case of $j=n$, e.t.

then combining with (11), we have (8). □

(2) for $0<p_1<1$, $p_i<0$ ($i=2,\dots ,n$), we have the reverse of (12).

Proof (1) For $p_i>1$ ($i=1,\dots ,n$), by the Hölder inequality (cf. [18]) and (7), it follows

(13)

Then by (7), we have (12). (Note: for $n=2$, we do not use the Hölder inequality again in the above.) (2) For $0<p_1<1$, $p_i<0$ ($i=2,\dots ,n$), by the reverse Hölder inequality and in the same way, we obtain the reverses of (12). □

With the assumptions given in Lemma 5, setting ${\varphi }_i(x):=x^{p_{{}_i(1-{\lambda }_i)-1}}$ ($x\in (0,\mathrm{\infty })$; $i=1,\dots ,n$), then we find ${\varphi }_n^{1/(1-p_n)}(x)=x^{q_{{}_n\lambda _n-1}}$. If $p_i>1$ ($i=1,\dots ,n$), define the following real function spaces:

If $f_i(\ge 0)\in L_{{\varphi }_i}^{p_i}(0,\mathrm{\infty })$, ${\parallel f\parallel }_{p_{i,{\varphi }_i}}>0$ ($i=1,\dots ,n$), then (i) for $p_i>1$ ($i=1,\dots ,n$), we have $\parallel T\parallel =k_{\lambda }$ and the following equivalent inequalities:

(19)

(20)

where the constant factor $k_{\lambda }$ is the best possible; (ii) for $0<p_1<1$, $p_i<0$ ($i=2,\dots ,n$), using the formal symbols in the case of (i), we have the equivalent reverses of (19) and (20) with the same best constant factor.

Proof (i) For all $p_i>1$, if (12) takes the form of equality, then for $n\ge 3$ in (14), there exist $C_i$ and $C_k$ ($i\ne k$) such that they are not all zero and

then $J={\{{\int }_0^{\mathrm{\infty }}{x}_n^{p_n(1-{\lambda }_n)-1}{f}_n^{p_n}(x_n)d{x}_n\}}^{\frac{1}{q_n}}$. By (12), it follows $J<\mathrm{\infty }$. If $J=0$, then (19) is naturally valid. Assuming that $0<J<\mathrm{\infty }$, by (20), it follows

and then (19) is valid, which is equivalent to (20).