Prove the following:
Theorem 1 If f (x, y) satisfies
def
∆f =
∂2f
∂2f
+
= 0,
∂x2
∂y 2
then the maximum of f over the rectangle
R = {(x, y) : 0 ≤ x ≤ 1, 0 ≤ x ≤ 1}
is achieved on the boundary of R.
Follow the following steps:
1. ( +5% Exam2 bonus) Assume ∆f > 0. Prove that there are no
local maxima inside the rectangle R.
2. ( +10% Exam2 bonus) Let ∆f = 0. Assume the contrary: there
is a local maximum point (x0 , y0 ) inside, such that f (x0 , y0 ) > M (f ),
where M (f ) is the
maximum of f over the boundary of R. Find an ε
so that g = f + ε (x − x0 )2 + (y − y0 )2 still satisfies g(x0 , y0 ) > M (g).
Conclude that g has a local maximum inside R. Use part 1 to obtain
a contradiction.
Your solution must be well explained yet succinct and, above all, logically correct. Do not confuse “necessary” with “sufficient” conditions.
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