# Analisi Due by Gianni Gilardi By Gianni Gilardi

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Example text

10. If G is noncompact, we have for ∞ > p1 > p2 ≥ 1 that MH Lp1 (G), p2 (H) = {0} . This also holds true for p1 = ∞, if we consider (weak*,weak)-continuous operators only. Proof. , see [9, Lem. 1 (i)]). Let T ∈ MH (Lp1 , T f + Th T f p2 = T (f + Th f ) p2 p2 Lp ). , Tf p2 p2 ≤ T 21/p1 f ≤ 21/p1 −1/p2 T f Lp1 Lp1 , Lp1 . 30 Georg Zimmermann for all f ∈ Lp1 (G). But since 21/p1 −1/p2 < 1, this contradicts the deﬁnition of T , unless T = 0 and thus T = 0. 5, T f = (f ∗ϕ)(h) h∈H for some ϕ ∈ L1 (G).

7). This completes the proof. 5. If f ∈ S(Rn ), then for 1 < p < n, the inequality f C ∇f p holds with q = np/(n − p). q ≤ Proof. 6) are satisﬁed with these weights. But this is clearly the case if q = p and 1/q = −1/n + 1/q . The condition n/(n − 1) < q is then equivalent to 1 < p < n and q = np/(n − p). This result was proved by G. Talenti  with a sharp constant. See also [4, Thm. 3]. 6) are satisﬁed only if p = q = 2 and n > 2. 1) that Rn |f (x)|2 |x|−2 dx 1/2 1/2 ≤C Rn |(∇f )(x)|2 dx , when n > 2.

By assumption, we have T (e2πi ηg ) = T (Mη 1) = Mη T (1) = e2πiηg ϕ for all η ∈ H ⊥ . Since trigonometric polynomials are dense in Lp1 (G/H), we have T = Qϕ . If p1 ≥ p2 < ∞, we have that 38 Georg Zimmermann A ϕ ∈ Lp2 (G) ⇐⇒ ∀A ∈ Lp1 (G/H) |A|p2 |ϕ|p2 ∈ L1 (G) ⇐⇒ |A|p2 |ϕ|p2 ⇐⇒ B |ϕ|p2 ◦ H ⇐⇒ ◦ H ∀A ∈ Lp1 (G/H) ∈ L1 (G/H) ∀A ∈ Lp1 (G/H) ∈ L1 (G/H) |ϕ|p2 ◦ H ∀B ∈ Lp1 /p2 (G/H) ∈ Lp1 /(p1 −p2 ) (G/H) , so Qϕ ∈ M·H ⊥ Lp1 , Lp2 if and only if ϕ ∈ Lp2 ,r (H, G/H). Then we can estimate for A ∈ Lp1 (G/H) that Qϕ A p2 Lp2 Aϕ = p2 dµG G A = p2 |ϕ|p2 G/H ≤ A (∗) G/H = A p2 p L 1 p1 ◦ H dµG/H p2 /p1 |ϕ|p2 dµG/H G/H ϕ p2 p ,r L 2 ◦ H r/p2 p2 /r dµG/H , where inequality (∗) follows from H¨ older’s inequality with dual exponents ◦ r/p2 p1 (p, q) = (p1 /p2 , r/p2 ).