In the sequel, we will suppose that Δ ≠ 0 with Δ defined by (2.3).
3.1 Sufficient conditions
Since, due to (2.6), we have that
it yields that
Hence: if we derive sufficient conditions for the two Hardy-type inequalities
(3.1)
(3.2)
we obviously obtain also sufficient conditions for the inequality (1.6) to hold.
Let us first consider (3.1). Due to (2.7), inequality (3.1) will be satisfied if there will be
(3.3)
for n = 1, 2,..., k. If we denote h(t) = g(t)tn-1, we can rewrite (3.3) as
(3.4)
But this is just the Hardy inequality for the function h with weight functions U(x) = |P
n
(x)|qu(x), V(x) = x-(n-1)pv(x), and it is well-known that this inequality holds for 1 < p ≤ q < ∞ if and only if the function
(3.5)
is bounded, while for the case 1 < q < p < ∞, the necessary and sufficient condition reads
(3.6)
here and in the sequel and (for details, see, e.g., [2].)
Now, let us consider (3.2). Analogously as in the foregoing case, (3.2) will be satisfied, if--due to (2.8)--the following Hardy-type inequality for the function h with weight functions U(x) = |Q
n
(x)|qu(x), V(x) = x-(n-1)pv(x) will be satisfied:
(3.7)
In this case, it is well-known (see, e.g., [4]) that the boundedness of the function
(3.8)
for 1 < p ≤ q < ∞ or the finiteness of the number
(3.9)
for 1 < q < p < ∞ is necessary and sufficient for (3.7) to hold.
Consequently, we have found sufficient conditions of the validity of the k-th order Hardy inequality (1.1):
Theorem 3.1. Let 1 < p, q < ∞ and for k ∈ ℕ, let n = 1, 2,..., k. Let P
n
(x) and Q
n
(x) be the polynomials from (2.7) and (2.8), respectively. Let AM,n(x) and be defined by (3.5) and (3.8), respectively, and BM,nand by (3.6) and (3.9), respectively. Then the k-th order Hardy inequality (1.1) holds for functions f satisfying the boundary conditions (1.2) if the weight functions u, v satisfy for n = 1, 2,..., k the conditions
(3.10)
in the case 1 < p ≤ q < ∞, and the conditions
(3.11)
in the case 1 < q < p < ∞.
3.2 Necessary and sufficient conditions
The Hardy inequality of higher order is, as we have seen, closely connected with the weighted norm inequality (1.6). This inequality with rather general kernels K(x, t) was investigated by many authors, see e.g. [2, 5]. Here, we use the fact that K(x, t) is a Green function and we assume that 1 < p < ∞, q > 0 and that
(3.12)
Let us denote Δ1 and Δ2 the closed triangles {(x, t): a ≤ t ≤ x ≤ b} and {(x, t): a ≤ x ≤ t ≤ b}, respectively. Due to (2.6), (2.7), and (2.8), we have that
(3.13)
Furthermore, suppose that
(3.14)
Theorem 3.2. Let 1 < p < ∞, q > 0 and suppose that (3.12), (3.13) and (3.14) hold. Then the Hardy-type inequality (1.6) holds if and only if
(3.15)
Proof. Necessity: Suppose that (1.6) holds.
(i) Due to (3.14), there exists a point t
a
∈ (a, b) such that G2(a, t
a
) ≠ 0. Consequently, there exists ε > 0 such that |G(x, t)| = |G2(x, t)| ≥ C
a
> 0 for all (x, t) ∈ (a, a+ε) × (t
a
- ε, t
a
+ε). Here we suppose that [t
a
- ε, t
a
+ ε] ⊂ (a, b). If we choose the test function as , we get from (1.6) that
i.e.,
due to (3.12). Together with (3.12), the last inequality implies that
which means that
(3.16)
(ii) Due to (3.14), there exists a point t
b
∈ (a, b) such that G1(b, t
b
) ≠ 0 and |G(x, t)| = |G2(x, t)| ≥ C
b
> 0 for all (x, t) ∈ (b-ε, b)×(t
b
-ε, t
b
+ε). The choice leads analogously as in (i) to the estimate
i.e.
so that
This together with (3.16) and (3.12) gives that u ∈ L1(a, b).
(iii) Due to (3.14), there exists a point x
a
∈ (a, b) such that G1(x
a
, a) ≠ 0 and for all (x, t) ∈ (x
a
- ε, x
a
+ ε) × (a, a + ε). Let us choose a test function in (1.6) as
where δ ∈ (0, ε) is a parameter. Then we get that
i.e., that
This estimate holds for all δ ∈ (0, ε), and with δ tending to zero on the left hand side of the estimate we obtain that
which implies that .
(iv) Finally, we obtain analogously from G2(x
b
, b) ≠ 0 that . Hence and the necessity is proved.
Sufficiency: Using the boundedness of the function G(x, t) (which follows from (3.13)), Holder's inequality and (3.15), we can estimate the left hand side of (1.6) as follows:
The proof is complete.
Remark 3.3. We have considered the Hardy-type inequality (1.6) for the case that G(x, t) was a Green function, i.e., G
i
(x, t) have been polynomials. It is obvious that we can repeat our approach for any function G(x, t), which satisfies (3.13) and (3.14). Hence, our approach gives some new criteria for the validity of (1.6) for rather general kernels G.
Example 3.4. In Example 1.1, the first order Hardy inequality with boundary condition (1.3) was considered. It can be easily shown that in this case the Green function has the form
where α + β ≠ 0. If α ≠ 0 and β ≠ 0, and then the conditions (3.14) are satisfied and we can use Theorem 3.2. According to this theorem, the Hardy inequality (1.6) holds if and only if
Example 3.5. For simplicity let us assume for (a, b) the interval (0, 1) and consider the second order Hardy inequality
(3.17)
Then boundary conditions (1.2) take the following form:
(3.18)
This inequality was considered in [4] and the corresponding Green function has the following form:
where
with λ
i
:= α
i
,1 + β
i
,1, μ
i
:= α
i
,1 + β
i
,1 + β
i
,2, νi := β
i
,1 + β
i
,2, i = 1, 2. Notice, that Δ is the corresponding determinant from (2.3).
Let us use Theorem 3.2; for this aim we consider the polynomials:
These polynomials satisfy conditions (3.14) if and only if
(3.19)
and these conditions imply that the second order Hardy inequality holds if and only if .
If the condition (3.14) is violated, then Theorem 3.2 cannot be used. Nevertheless, in some cases, it is possible to use the following generalization:
Theorem 3.6. Suppose that 1 < p < ∞, q > 0 and the functions G
i
(x, t) (i = 1, 2) are not identically equal to zero.
(i) If the Hardy-type inequality (1.6) holds, then there exist polynomials P
i
(x), Q
i
(t) (i = 1, 2) on (a, b) such that
(3.20)
and that the corresponding Green function G(x, t) can be written as
(3.21)
where the functions satisfy (3.14).
If, moreover, , then
(i-1) for p ≤ q
(3.22)
where
(3.23)
(3.24)
(i-2) for q < p
(3.25)
where
(3.26)
(3.27)
(ii) If there exist polynomials P
i
(x), Q
i
(t) on (a, b) (i = 1, 2) such that (3.21) holds and the conditions (3.22) (for p ≤ q), (3.25) (for q < p) are satisfied, then the Hardy-type inequality (1.6) holds.
Proof. (i) Let the Hardy-type inequality (1.6) hold, then the following inequality
(3.28)
also holds for arbitrary function f ∈ Lp(v), which follows from (1.6) considered for the function and then from the monotonicity of the outer integral on the left hand side of (1.6).
(i.1) If G2(a, t) does not vanish identically on (a, b), then the proof of the existence of the polynomial P1(t) follows from point (i) of the proof of Theorem 3.2, i.e., in this case and the polynomial can be chosen as P2(x) ≡ 1.
(i.2) If G2(a, t) vanishes on (a, b), then there exists a positive integer α2 such that , where does not vanish on (a, b). Choosing ε > 0 in inequality (3.28) sufficiently small and repeating the calculations in point (i) of the proof of Theorem 3.2 we obtain that
which implies that and the polynomial can be chosen as .
(i.3) Similarly, we can prove that there exist nonnegative integers α1, β1, β2 such that
and the polynomials can be chosen as
Moreover, it can be easily shown that the weight functions with these polynomials satisfy (3.20) and (3.21).
(i.4) Now we show that the conditions (3.22) and (3.25) are satisfied. Using (3.21) we rewrite (3.28) in the form
and taking into account that (i = 1,2) we obtain the following equivalent inequality
for all f ∈ Lp(v) and for sufficiently small ε > 0. Using Theorem 2.3 in [2] we obtain the following equivalent conditions on the interval (a, a + ε)
(for p ≤ q)
(3.29)
(for q < p)
(3.30)
Similarly, we obtain the following conditions on the interval (b - ε, b):
(for p ≤ q)
(3.31)
(for q < p)
(3.32)
All these conditions together with (3.20) imply that conditions (3.22) and (3.25) are satisfied:
Let us prove (i-1). Using (3.20) it is easy to show that the condition is satisfied if and only if there exist the limits
Otherwise, the existence of these limits is equivalent to the existence of
For the proof of this assertion, we only show the following equality, since the others can be proved analogously:
The existence of the limits follows from (3.29) and (3.31) and (i-1) is obtained.
To prove (i-2) is enough to show B1(a, b) < ∞, since the case B2(a, b) < ∞ can be proved analogously. First we rewrite B1(a, b) in the form
The boundedness of I2 follows from (3.20). Moreover, (3.20) together with (3.30) implies that
and
To get the last estimate we used that
which follows from (3.20).
Finally, we obtain (ii) only using boundedness of the polynomials and [2, Theorem 2.3]. The proof is, then, complete.
Example 3.7. Let us go back to Example 3.5. If, e.g., |a| + |b| = 0, then one of the conditions (3.19) is violated. In this case we proceed according to Theorem 3.6 where , and ; if, moreover, c + Δ = 0, then where , and the Hardy inequality (3.17) holds for functions satisfying (3.18) if and only if (3.22) (for p ≤ q) or (3.25) (for q < p) hold with P1(x) = c + dx, Q1(t) = t; P2(x) = x, Q2(t) = dt - Δ. The other cases of violation of (3.19) can be considered analogously.