A New Extension of Serrin′s Lower Semicontinuity Theorem

Theorem 1 (see [3].)In addition to (H1), if f satisfies one of the following conditions:

• (a)

  f(x, s, ξ)→+∞ when |ξ | →+∞, for all (x, s) ∈ Ω × R,

• (b)

  f(x, s, ξ) is strictly convex in ξ ∈ Rⁿ for all (x, s) ∈ Ω × R,

• (c)

  the derivatives f_x(x, s, ξ), f_ξ(x, s, ξ), and f_ξx(x, s, ξ) exist and are continuous for all (x, s, ξ) ∈ Ω × R × Rⁿ.

Theorem 2 (see [8], [9].)Let one assume that f satisfies (H1) and that, for every compact set K ⊂ Ω × R × Rⁿ, there exists a constant L = L(K) such that

and, for every compact set K₁ ⊂ Ω × R, there exists a constant L₁ = L₁(K₁) such that Then F(u, Ω) is lower semicontinuous in $$ with respect to the strong convergence in $$.

Theorem 3 (see [8], [9].)Let f satisfy (H1) such that, for every open set Ω^′ × H × K ⊂ ⊂Ω × R × Rⁿ, there exists a constant $$ such that, for every x₁, x₂ ∈ Ω^′, s ∈ H, and ξ ∈ K,

Then the functional F(u, Ω) is lower semicontinuous in $$ with respect to the $$ convergence.

Theorem 4. Let Ω ⊂ R be an open set; f(x, s, ξ) : Ω × R × R → [0, +∞) satisfies the following conditions:

• (H1)

   f(x, s, ξ) is continuous on Ω × R × R, and, f(x, s, ξ) is convex in ξ ∈ R for all (x, s) ∈ Ω × R;

• (H2) 

  f_ξ(x, s, ξ) is continuous on Ω × R × R, and for every compact set of Ω × R × R, f_ξ(x, s, ξ) is absolutely continuous about x;

• (H3)

   for every compact set K₁ ⊂ Ω × R, there exists a constant L₁ = L₁(K₁), such that

  $$ ()
  $$ ()

Then the functional F(u, Ω) = ∫_Ω f(x, u(x), u^′(x))dx is lower semicontinuous in $$ with respect to the strong convergence in $$.

Theorem 5. Let Ω ⊂ R be an open set; f(x, s, ξ) : Ω × R × R → [0, +∞) satisfies (H1) and the following condition:

• (H4)

   for every compact set of Ω × R × R, f(x, s, ξ) is absolutely continuous about x.

Then the functional F(u, Ω) is lower semicontinuous in $$ with respect to the strong convergence in $$.

Lemma 6 (see [10].)Let U⊆R^d be an open set and f : U × Rⁿ → [0, +∞) a continuous function with compact support on U, such that, for every t ∈ U, f(t, ·) is convex on Rⁿ. Then there exists a sequence $$, and supp (η_q)⊆B(0,1), such that, if we let

one has satisfying the following results:

• (i)

  for every j ∈ N, f_j : U × Rⁿ → [0, +∞) is a continuous function with compact support on U such that, for all  t ∈ U, f_j(t, ·) is convex on Rⁿ. Moreover, for all (t, ξ) ∈ U × Rⁿ,   f_j(t, ξ) ≤ f_j+1(t, ξ), and

  $$ ()

• (ii)

  for every j ∈ N, there exists a constant M_j > 0, such that, for all (t, ξ) ∈ U × Rⁿ,

  $$ ()
  and, for all t ∈ U, and ξ₁, ξ₂ ∈ Rⁿ;
  $$ ()

Step 1. Let {β_i(x, s)} _i∈N be a sequence of smooth functions satisfying

• (1)

  there exists a compact set Ω^′ × H ⊂ ⊂Ω × R, such that β_i(x, s) = 0, for all (x, s) ∈ (Ω∖Ω^′) × (R∖H);

• (2)

  for every i ∈ N,   β_i(x, s) ≤ β_i+1(x, s), for all (x, s) ∈ Ω^′ × H;

• (3)

  lim _i→+∞β_i(x, s) = 1, for all (x, s) ∈ Ω^′ × H.

Step 2. Let $$ be a mollifier, and, for ϵ > 0, define

where [Ω_ε]≜{x ∈ Ω : dist (x, ∂Ω) > ε}. We have In the following, we denote the derivative of u_ε by $$. When $$, we know $$. By the properties of convolution, we know $$ and That is, for  all  δ > 0,   ∃ ϵ > 0, such that

Step 3. Now, we estimate the right side of inequality (27).

By (6) and (24), we have

Thus Since f(x, s, ξ) and f_ξ(x, s, ξ) are continuous functions, they are bounded functions on compact subset. By Lebesgue Dominated Convergence Theorem, we obtain Now, we will prove By Lemma 6, there exists a sequence of nonnegative continuous functions f_j(x, s, ξ)    (j ∈ N), such that f_j(x, s, ξ) is convex on ξ, and, for all (x, s, ξ) ∈ Ω^′ × H × R, By Levi’s Lemma, we obtain In order to prove (31), we only need to prove By (33), we have Thus (31) holds.

Step 4. Now, we need to prove

Let By triangle inequality and (7), we have By (39), condition (H2) and $$, we know that g(x, s) is a locally absolute continuous function about x. So g(x, s) is almost everywhere differentiable; that is, ∂g/∂x exists almost everywhere. Taking derivatives in both sides of (38), we obtain Because G_m(x) vanishes outside Ω^′, we obtain By (40), we have where We note By Fubini’s Theorem, we have Since g(x, s) is absolutely continuous about x, ∂g/∂x is integrable and absolutely integrable with respect to x; that is, By (17) and (46), we obtain Because of the absolute continuity of integral, we have By (42), we obtain Thus we just proved (36). By (29)–(31) and (36), we have Thus we deduce that the functional F(u, Ω) is lower semicontinuous in $$ with respect to the strong convergence in $$. We complete the proof.

Step 1. Similar to the first step of the proof in Theorem 4, without loss of generality, we assume that the integrand f(x, s, ξ) vanishes outside a compact subset of Ω × R. Thus we assume that there exists an open set Ω^′ × H ⊂ ⊂Ω × R, such that

Let $$, such that u_m → u in $$; we need to prove

By Lemma 6, there exists a function sequence {f_j(x, s, ξ)} _j∈N, such that, for all  j ∈ N, f_j is a continuous function on Ω^′ × H ⊂ ⊂Ω × R, for all (x, s) ∈ Ω^′ × H,  f_j(x, s, ·) is convex on R, and, for  all  (x, s, ξ) ∈ Ω^′ × H × R, Let $$ be a mollifier, and define the f_j,ε = f_j*η_ε; that is, By (55), we have Choose ε = ε_j = 1/jM_j → 0. By (57), we have So By (53), (54), and Levi’s Lemma, we have Let By (59)–(61), we have Obviously, Thus Therefore F(u, Ω^′), being the supremum of the family of functionals {F_j(u, Ω^′)} _j∈N, will be lower semicontinuous if every {F_j(u, Ω^′)} is lower semicontinuous.

Step 2. In order to prove that, for all  j ∈ N, F_j(u, Ω^′) is lower semicontinuous in $$ with respect to the strong convergence in $$, we will prove that, for all  j ∈ N, the integrand $$ satisfies all conditions of Theorem 4. Obviously, for  all  j ∈ N, $$ satisfies condition (H1).

For all (x, s) ∈ Ω^′ × H and ξ₁, ξ₂ ∈ R, by (55), we have

Thus So $$ satisfies (6) in condition (H3) of Theorem 4.

Step 3. Next we will prove that $$ satisfies condition (H2).