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Chapter 1: Preliminaries                                          175

                                                     R
                Exercise 1.3.6. (i) Let f ∈ C 0 ∞  (Ω) with  Ω  f (x) dx =1,bea fixed function.
                Let w ∈ C 0 ∞  (Ω) be arbitrary and
                                                 ∙Z        ¸
                                   ψ (x)= w (x) −   w (y) dy f (x) .
                                                   Ω
                                               R
                We therefore have ψ ∈ C ∞  (Ω) and  ψ =0 which leads to
                                     0
                        Z                Z                Z              Z
                 0=       u (x) ψ (x) dx =  u (x) w (x) dx −  f (x) u (x) dx ·  w (y) dy
                         Ω                Ω                Ω               Ω
                        Z ∙       Z             ¸
                    =      u (x) −   u (y) f (y) dy w (x) dx .
                         Ω         Ω
                                                              R
                Appealing to Theorem 1.24, we deduce that u (x)=  u (y) f (y) dy = constant
                a.e.
                                           R  b
                   (ii) Let ψ ∈ C 0 ∞  (a, b),with  a  ψ =0, be arbitrary and define
                                                Z  x
                                         ϕ (x)=     ψ (t) dt .
                                                  a
                Note that ψ = ϕ and ϕ ∈ C  ∞  (a, b). We may thus apply (i) and get the result.
                              0
                                        0
                Exercise 1.3.7. Let, for ν ∈ N,
                                       u ν (x)= min {|u (x)| ,ν} .
                The monotone convergence theorem implies that, for every  > 0,wecan find ν
                sufficiently large so that
                                    Z            Z

                                      |u (x)| dx ≤  u ν (x) dx +  .
                                     Ω            Ω            2
                Choose then δ =  /2ν. We therefore deduce that, if meas E ≤ δ,then
                  Z            Z             Z

                     |u (x)| dx =  u ν (x) dx +  [|u (x)| − u ν (x)] dx ≤ ν meas E +  ≤  .
                   E            E             E                              2
                For a more general setting see, for example, Theorem 5.18 in De Barra [37].

                7.1.3   Sobolev spaces

                Exercise 1.4.1. Let σ n−1 =meas (∂B 1 (0)) (i.e. σ 1 =2π, σ 2 =4π,...).
                   (i) The result follows from the following observation

                                    Z                   Z  R
                               p             p              n−1      p
                            kuk L p =   |u (x)| dx = σ n−1  r   |f (r)| dr.
                                                         0
                                     B R
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