An Opial-Type Inequality on Time Scales

Theorem A (see  [1]). If f ∈ C¹[0, h] satisfies f(0) = f(h) = 0 and f(x) > 0 for all x ∈ (0,   h), then

Theorem B (see  [12]). If f is absolutely continuous on [0, h] with f(0) = 0, then

The equality in (2) holds if and only if f(x) = cx, where c is a constant.

Theorem C (see  [10]). Let x(t) ∈ C⁽ⁿ⁾[0, a] be such that x⁽ⁱ⁾(0) = 0,   0 ≤ i ≤ n − 1  (n  ≥   1). Then the following inequality holds:

Lemma 1 (see  [16]). Let a  ≥   0,   p  ≥ 1 be real constants. Then

for any k > 0.

Lemma 2 (see [17].)Let $$ the set of functions that are m times differentiable with rd-continuous derivatives on 𝕋,  m ∈ ℕ. Then for any a,  b ∈ 𝕋 and t ∈ [a, b]∩𝕋,

where $$.

Theorem 3. Let 0 ≤ r ≤ s < t, s > 0, t > 1 be real numbers, and let m, k be integers with 0 ≤ k ≤ m − 1. Let p > 0 and q ≥ 0 be measurable functions on Υ : = [a, b]∩𝕋. Further, let $$ with $$, and let$$ be absolutely continuous on Υ such that the integrals $$and $$exist. Then one has

where α : = s/t, β : = r/s,

Proof. Since $$, we obtain from Taylor′s theorem that for all x ∈ Υ,

and hence

From (9) and Hölder′s inequality we get

where $$,  $$.

Let

Then we have

So (10) together with (12) implies

where h(x): = q(x)p(x) ^−s/tP(x) ^r(t−1)/t. Integrating both sides of (13) over Υ and making use of Hölder′s inequality, we obtain

Similarly, we get

Recall the elementary inequalities

where

Let β = r/s. Since α = s/t ∈ (0, 1) and F is nondecreasing, from (14)–(16), we have

By Lemma 1, we get

From (18) and (19), we conclude

The proof is complete.

Theorem 4. Let r ≥ 0, s > 0, s < t, t > 1 be real numbers, and let m, k be integers with 0 ≤ k ≤ m − 1. Let p > 0, and q ≥ 0 be measurable functions on Υ : = [a,   b]∩𝕋. Further, let f, $$ with let $$, and $$ be absolutely continuous on Υ such that the integrals $$and $$exist. Then one has

where $$, β : = r/s, α : = s/t, h(x): = q(x)p(x) ^−αP(x) ^r(t−1)/t, $$, and

Proof. Following the proof of Theorem 3, we obtain

Using (16),

The proof is complete.

Remark 5. In the special case where 𝕋 = ℝ, Theorem 4 reduces to Theorem 1 of [13].

Theorem 6. Let $$, Υ : = [a,   b]∩𝕋 be such that $$, 0 ≤ k ≤ m − 1, let $$ be absolutely continuous on Υ, and let $$. Then

Proof. From the hypotheses, we have

Multiplying (26) by $$ and using Cauchy-Schwarz inequality, we obtain

Integrating both sides over x from a to b and using Cauchy-Schwarz inequality, we observe

The proof is complete.

Theorem 7. Let p(x) > 0, q(x) be nonnegative and measurable on Υ = [a, b]∩𝕋, and let $$ be such that $$, 0 ≤ k ≤ m − 1. If $$ is absolutely continuous on Υ, then for r > 1, r_k > 0, and any 0 ≤ r_m < r,

where $$, $$,$$.

Proof. Following the hypotheses, it is easy to see that (26) holds. By using Hölder′s inequality with indices r and r/(r − 1), we obtain

where $$, $$. So we get and hence for any r_m,

Thus for r_k > 0,

Integrating both sides of (33) from a to b and applying Hölder′s inequality with indices r/r_m and r/(r − r_m), we obtain

where $$. The proof is complete.

Remark 8. In the special case where 𝕋 = ℝ, Theorems 6 and 7 reduce to Theorems 1 and 2 of [18].