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In addition, G σ (x, ·)−c is nonnegative superharmonic, and representable because the sum of u x and a harmonic functionality. due to the fact that G σ (x, ·) − c < G σ (x, ·), the minimality ∇ of G σ (x, ·) is contradicted. Lemma four. 2. 6 If the open set σ has a eco-friendly functionality and B is a ball with x ≥ B ∞ B − ∞ σ, then G σ (x, ·) is bounded outdoors B. evidence G σ (x, ·) is continuing and nonnegative on σ ∼ B. It has to be strictly confident at the element of σ containing x, for if G σ (x, y) = zero for a few aspect y during this part, then G σ (x, ·) will be identically 0 thereon by way of the minimal precept, Corollary three. three. 6, in contradiction to the truth that G σ (x, x) = +⊂. because B − ∞ σ, B − is a subset of the part containing x and G σ (x, ·) is precisely optimistic on B − . enable m = supy≥π B G σ (x, y) > zero. outline g(x, y) = G σ (x, y), min (G σ (x, y), 2m), y ≥ B, y ≥ σ ∼ B. Then g(x, ·) is superharmonic on B because it concurs with G σ (x, ·) on B. for the reason that g(x, ·) is the minimal of 2 superharmonic services on σ ∼ B − , it truly is superharmonic on σ ∼ B − . seeing that G σ (x, ·) ≤ m < 2m on π B and G σ (x, ·) is continuing at issues of π B, g(x, ·) is of the same opinion with G σ (x, ·) on a local of π B and is superharmonic thereon. It follows that g(x, ·) is superharmonic on σ. in view that g(x, ·)−u x = G σ (x, ·)− u x on B ∼ {x} and G σ (x, ·) − u x is harmonic on B, g(x, ·) − u x could be outlined at x to be superharmonic on σ. Denoting the extension by means of [g(x, ·) − u x ], 4. 2 eco-friendly capabilities 127 zero ≤ g(x, ·) = u x + [g(x, ·) − u x ], the place the bracket functionality is superharmonic on σ and G σ (x, ·) ≤ g(x, ·) through the definition of the golf green functionality. yet on the grounds that g(x, ·) ≤ 2m on σ ∼ B, G σ (x, ·) ≤ 2m on σ ∼ B. ∇ The lifestyles of a eco-friendly functionality for an open set σ might be tested via exhibiting that σ helps sufficiently many nonnegative superharmonic capabilities. Definition four. 2. 7 If σ is an open subset of R n and x ≥ σ, permit Bx = {vx ; vx ≤ zero, vx = u x + wx , wx ≥ S(σ)}. the category Bx will more often than not depend on σ and x. it may possibly be that Bx = ∅ for a few x ≥ σ. Lemma four. 2. eight Bx is a saturated relatives of superharmonic capabilities on σ ∼ {x}. facts because there's not anything to end up if Bx = ∅, it may be assumed that Bx →= ∅. If v1 , v2 ≥ Bx , then vi ≤ zero and vi = u x + wi , i = 1, 2, the place w1 and w2 are superharmonic on σ. because min (v1 , v2 ) = u x + min (w1 , w2 ) and min (w1 , w2 ) is superharmonic on σ, min (v1 , v2 ) is nonnegative and superharmonic on σ; that's, the minimal of 2 components of Bx belongs to Bx . believe now that v = u x + w ≥ Bx the place w is superharmonic on σ. enable B be any ball with B − ∞σ∼{x}. via Lemma three. four. eleven, vB = PI(v : B) v on B on σ ∼ B is superharmonic on σ. seeing that u x is harmonic on an area of B − , PI(v : B) = PI(u x + w : B) = u x + PI(w : B). therefore, v B = u x + w B is superharmonic on σ ∇ and v B ≥ Bx . This completes the facts that Bx is saturated on σ ∼ {x}. Lemma four. 2. nine If Bx →= ∅ for every x ≥ σ, then σ has a eco-friendly functionality G σ (x, ·) = inf {vx ; vx ≥ Bx }. facts contemplate a set x ≥ σ and enable wx = inf {vx ; vx ≥ Bx }.