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**Sample text**

15) i,j=1 for some ν > 0. Moreover, if Ω = RN we need that the leading coefficients aij are uniformly continuous. The following results hold. 1 (S. Agmon, [1]) Let p ∈ (1, +∞). (i) Let Ap : W 2,p (RN ) → Lp (RN ) be defined by Ap u = Au. The operator Ap is sectorial in Lp (RN ) and D(Ap ) is dense in Lp (RN ). (ii) Let Ω and A be as above, and let Ap be defined by D(Ap ) = W 2,p (Ω) ∩ W01,p (Ω), Ap u = Au. Then, the operator Ap is sectorial in Lp (Ω), and D(Ap ) is dense in Lp (Ω). 16) i=1 the coefficients bi , i = 1, .

1) u(0) = x, where f : [0, T ] → X. Throughout the chapter we use standard notation. We recall that if Y is any Banach space and a < b ∈ R, B([a, b]; Y ) and C([a, b]; Y ) are the Banach spaces of all bounded (respectively, continuous) functions from [a, b] to Y , endowed with the sup norm f ∞ = supa≤s≤b f (s) Y . C α ([a, b]; Y ) is the Banach space of all α-H¨older continuous functions from [a, b] to Y , endowed with the norm f C α ([a,b];Y ) = f ∞ + [f ]C α ([a,b];Y ) , where [f ]C α ([a,b];Y ) = supa≤s,t≤b f (t) − f (s) Y /(t − s)α .

Since ∂Gt /∂t = ∆Gt , then u solves the heat equation in (0, +∞) × RN . Let us show that T (t)f → f in X as t → 0+ if f ∈ Lp (RN ) or f ∈ BU C(RN ). If f ∈ Lp (RN ) we have T (t)f − f p p p Gt (y)f (x − y)dy − f (x) dx = RN RN p Gt (y)[f (x − y) − f (x)]dy dx = RN RN G1 (v)[f (x − = RN G1 (v)|f (x − = p tv) − f (x)]dv dx RN ≤ RN √ √ tv) − f (x)|p dv dx RN |f (x − G1 (v) RN √ tv) − f (x)|p dx dv. RN Here we used twice the property that the integral of Gt is 1; the first one to put f (x) under the integral sign and the second one to get G1 (v)[f (x − √ tv) − f (x)]dv p ≤ RN G1 (v)|f (x − √ tv) − f (x)|p dv RN √ through H¨older inequality, if p > 1.