12.5. MOLLIFIERS AND DENSITY OF SMOOTH FUNCTIONS 367

Proof: Consider the difference quotient for calculating a partial derivative of f ∗g.

f ∗g(x+ te j)− f ∗g(x)t

=∫

f (y)g(x+ te j−y)−g(x−y)

tdµ (y) .

Using the fact that g ∈ C∞c (Rn), the quotient g(x+te j−y)−g(x−y)

t is uniformly bounded.To see this easily, use Theorem 7.5.2 on Page 190 to get the existence of a constant, Mdepending on max{||Dg(x)|| : x ∈ Rn} such that

∣∣g(x+ te j−y)−g(x−y)∣∣ ≤M |t| for

any choice of x and y. Therefore, there exists a dominating function for the integrand ofthe above integral which is of the form C | f (y)|XK where K is a compact set dependingon the support of g. It follows the limit of the difference quotient above passes inside theintegral as t → 0 and ∂

∂x j( f ∗g)(x) =

∫f (y) ∂

∂x jg(x−y)dµ (y) . Now letting ∂

∂x jg play

the role of g in the above argument, partial derivatives of all orders exist. A similar use ofthe dominated convergence theorem shows all these partial derivatives are also continuous.

For the last claim, it is clear that spt( f ∗ψm)⊆ spt( f )+B(0,1/m) since off spt( f )+B(0,1/m) the integral for f ∗ψm will be 0. To verify the last claim, let ε > 0 be given. Byuniform continuity of f , | f (x)− f (x−y)|< ε whenever |y| is sufficiently small. There-fore,

| f (x)− f ∗ψm (x)| =

∣∣∣∣∫ ( f (x)− f (x−y))ψm (y)dµ (y)∣∣∣∣

≤∫

B(0,1/m)| f (x)− f (x−y)|ψm (y)dµ (y)< ε

∫ψmdµ = ε

whenever m is large enough. ■

Theorem 12.5.8 For each p≥ 1, C∞c (Rn) is dense in Lp(Rn). Here the measure is

Lebesgue measure.

Proof: Let f ∈ Lp(Rn) and let ε > 0 be given. Choose g∈Cc(Rn) such that ∥ f −g∥p <ε

2 . This can be done by using Theorem 12.2.4. Now let

gm (x) = g∗ψm (x)≡∫

g(x−y)ψm (y)dmn (y) =∫

g(y)ψm (x−y)dmn (y)

where {ψm} is a mollifier. It follows from Lemma 12.5.7 gm ∈ C∞c (Rn). It vanishes if

x /∈ spt(g)+B(0, 1m ).

∥g−gm∥p =

(∫|g(x)−

∫g(x−y)ψm(y)dmn(y)|pdmn(x)

) 1p

≤(∫

(∫|g(x)−g(x−y)|ψm(y)dmn(y))

pdmn(x)

) 1p

≤∫ (∫

|g(x)−g(x−y)|pdmn(x)

) 1p

ψm(y)dmn(y)

=∫

B(0, 1m )∥g−gy∥pψm(y)dmn(y)<

ε

2

whenever m is large enough thanks to the uniform continuity of g. Theorem 12.1.11 wasused to obtain the third inequality. There is no measurability problem because the function

(x,y)→ |g(x)−g(x−y)|ψm(y)

12.5. MOLLIFIERS AND DENSITY OF SMOOTH FUNCTIONS 367Proof: Consider the difference oe for calculating a partial derivative of f * g.fxg(a+te;)—fxg(x g(a+te;—y)—g(@—-y)t = [ry tdu(y).g(a+te;—y)—g(@—y)Using the fact that g € C?(R"), the quotient ; is uniformly bounded.To see this easily, use Theorem 7.5.2 on Page 190 to get the existence of a constant, Mdepending on max {||Dg (a)|| : @ € R"} such that |g(a+te;—y) —g(a—y)| <M|t| forany choice of x and y. Therefore, there exists a dominating function for the integrand ofthe above integral which is of the form C|f (y)| 2x where K is a compact set dependingon the support of g. It follows the limit of the difference quotient above passes inside theintegral as t + 0 and aa 7 (f #8) (@ J=ffiy )e g(a —y)du(y). Now letting 2 on 78 Playthe role of g in the above argument, partial den ives of all orders exist. A similar use ofthe dominated convergence theorem shows all these partial derivatives are also continuous.For the last claim, it is clear that spt (f « y,,,) C spt(f) + B(0,1/m) since off spt (f) +B(0,1/m) the integral for f * y,,, will be 0. To verify the last claim, let € > 0 be given. Byuniform continuity of f,|f (a) — f (a —y)| < € whenever |y| is sufficiently small. There-fore,If (@) — f* Vn (@)||e) raw) wwe)ros if (a) — f(@—Y)| Wn (y) aE (y) < € | vndu =€lAwhenever m is large enough. MfTheorem 12.5.8 For each p >1, Ce? (R") is dense in L?(IR"). Here the measure isLebesgue measure.Proof: Let f € L’(R”) and let € > 0 be given. Choose g € C,(R”) such that || f — g||p <5. This can be done by using Theorem 12.2.4. Now letim () = 8* Vy () = [ 8(@—y) Yn (y) dm») = | 8(y) Wu (@—¥) dmg (9)where {y,,} is a mollifier. It follows from Lemma 12.5.7 gm, € C?(R"). It vanishes ifx ¢ spt(g) + B(0, ;;)-le—enlle = (flee) [ve wsosnirnis)(f< [is@)-8@—W)l¥aly sg(u)Pay(a))/ (fist \e(@) — g(@—y) dmg (a 1) vata)= fey le Bull Ynlwdrmy) <in)IA IANl Mmwhenever m is large enough thanks to the uniform continuity of g. Theorem 12.1.11 wasused to obtain the third inequality. There is no measurability problem because the function(x,y) — |g(x) —9(@—y)|Vn(y)