Final exam solutions

# Solution 1 we have f x lim f n x x x 1 x 1 2 the

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Solution 1. We have f ( x ) = lim f n ( x ) = ± x, x < 1 0 , x = 1 2. The limit function f ( x ) is not continuous, hence the convergence is not uniform. 3. On [0 ,a ], the function f ( x ) - f n ( x ) = x n increases from 0 to a n . Hence lim sup x [0 , 1 / 2] | f n ( x ) - f ( x ) | = lim a n = 0 , so the convergence is uniform. 6

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Problem 7. 30 points. Suppose that f ( x ) is a continuous function on [ a,b ], suppose that f ( x ) 0 for all x [ a,b ], and suppose that Z b a f ( x ) dx = 0 . Prove that f ( x ) = 0 for all x [ a,b ]. (Hint: suppose that f ( c ) 6 = 0 for some c [ a,b ], and draw a picture). Proof. Suppose that f ( c ) > 0 for some c [ a,b ]. Then for any ε > 0 there exists a δ > 0 such that | x - c | < δ,x [ a,b ] implies | f ( x ) - f ( c ) | < ε. If we choose ε = f ( c ) / 2, then we ﬁnd that there is a δ > 0 such that | x - c | < δ and x [ a,b ] implies that f ( x ) > f ( c ) / 2. At least half of the interval [ c - δ,c + δ ] lies inside [ a,b ], and on that interval we have f ( x ) > f ( c ) / 2, hence the integral of f ( x ) over the part of the interval [ c - δ,c + δ ] lying inside [ a,b ] is greater than or equal to δ · f ( c ) / 2 > 0, which contradicts the assumption. 7
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Solution 1 We have f x lim f n x x x 1 x 1 2 The limit...

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