11.8. THE DIRICHLET PROBLEM IN GENERAL 283

Proof: I need to show that if xn → x, then f (x) ≤ liminfn→∞ f (xn). If not, thenthere exists, for some ε > 0 a subsequence, still denoted by n such that limn→∞ f (xn) <f (x)− ε. But then there is fk such that fk (x) > f (x)− ε

2 and so by continuity of fk,fk (x) = limn→∞ fk (xn)≤ limn→∞ f (xn)< fk (x)− ε

2 , a contradiction. ■This lemma is about finitely many subharmonic functions.

Lemma 11.8.2 Let U be an open set and let u1,u2, · · · ,up be subharmonic functionsdefined on U. Then letting v≡max(u1,u2, · · · ,up) , it follows that v is also subharmonic.

Proof: Let x ∈U. Then whenever r is small enough to satisfy the subharmonic condi-tion for each ui.

v(x) = max(u1 (x) ,u2 (x) , · · · ,up (x))

≤ max(

1ωn−1rn−1

∫∂B(x,r)

u1 (y)dσ (y) , · · · , 1ωn−1rn−1

∫∂B(x,r)

up (y)dσ (y))

≤ 1ωn−1rn−1

∫∂B(x,r)

max(u1,u2, · · · ,up)(y)dσ (y) =1

ωn−1rn−1

∫∂B(x,r)

v(y)dσ (y) .

This proves the lemma. ■

Definition 11.8.3 Let U be an open set and let u be subharmonic on U. Then forB(x0,r)⊆U define

ux0,r (x)≡

{u(x) if x /∈ B(x0,r)

1ωn−1r

∫∂B(x0,r) u(y) r2−|x−x0|2

|y−x|n dσ (y) if x ∈ B(x0,r)

Thus ux0,r is harmonic on B(x0,r) , and equals to u off B(x0,r) . This is because thereexists a harmonic function w whose boundary values on ∂B(x0,r) are given by u(y) and itequals ux0,r at x∈B(x0,r). The wonderful thing about this is that ux0,r is still subharmonicon all of U . Also note that, from Corollary 11.7.6 on Page 282, every harmonic function issubharmonic.

Lemma 11.8.4 Let U be an open set and B(x0,r)⊆U as in the above definition whereu is subharmonic. Then ux0,r is subharmonic on U and u≤ ux0,r.

Proof: First I show that u ≤ ux0,r. This follows from the maximum principle. Indeed,the function u−ux0,r is subharmonic on B(x0,r) and equals zero on ∂B(x0,r) . Thus forall ρ small enough,

u(z)−ux0r (z)≤1

ωρn−1

∫∂B(z,ρ)

u(y)−ux0,r (y)dσ (y)

thanks to the mean value property of harmonic functions, Corollary 11.7.6. Since this istrue for all small ρ , it follows from continuity that u(z)−ux0,r (z)≤ 0.The two functionsare equal off B(x0,r) . Thus for such z,

ux0,r (z) = u(z)≤ 1ωρn−1

∫∂B(z,ρ)

u(y)dσ (y)≤ 1ωρn−1

∫∂B(z,ρ)

ux0,r (y)dσ (y)

The second inequality is because there may be points of B(x0,r) in ∂B(z,ρ) if z ∈∂B(x0,r). ■