Note what about the converse example if ˆ f n c n 1

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Note. What about the converse? Example. If ˆ f ( n ) c n 1+ " , then ˆ f ( n ) 2 ` 1 , so f 2 A ( T ), and thus f is continuous and S n f ! f uniformly. Example. If ˆ f ( n ) c n 2+ " , then f is di erentiable. [See assignment] Lecture 33: July 22 3.12 Functions of Bounded Variation Definition. f : [ a, b ] ! R is ob bounded variation (BV) if there exists M such that for any partition P : a = x 0 < x 1 < · · · < x n = b , we have P n k =1 | f ( x k ) - f ( x k - 1 ) | . Example. 1. If f is increasing then P | f ( x k ) - f ( x k - 1 ) | = P f ( x k ) - f ( x k - 1 ) = f ( b ) - f ( a ), so f is of bounded variation. 2. If f is continuous on [ a, b ] and di erentiable on ( a, b ), and f 0 is bounded, then f is of BV. Note that this includes C 1 functions and twice di erentiable functions. Proof. Take any partition P . By the IVT, f ( x k ) - f ( x k - 1 ) x k - x k - 1 = f 0 ( t k ) for some t k 2 ( x k - 1 , x k ), so X | f ( x k ) - f ( x k - 1 ) | = X | f 0 ( t k ) || x k - x k - 1 |  k f 0 k 1 X ( x k - x k - 1 ) = k f 0 k 1 ( b - a ) .
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3 FOURIER ANALYSIS 52 3. y = ( sin 1 x x 6 = 0 c x = 0 is not of bounded variation on any interval containing 0. Notation. Let V f ( a, b ) = sup P P | f ( x k ) - f ( x k - 1 ) | denote the total variation of f over [ a, b ] . Proposition. If f is of bounded variation, then f is bounded. Proof. Consider the partition { a, x, b } . We have that | f ( x ) - f ( a ) | + | f ( b ) - f ( x ) | V f ( a, b ). Thus | f ( x ) | | f ( a ) | + V f ( a, b ). Proposition. If f and g are of bounded variation, then so are f ± g . Proof. Exercise – Use triangle inequality. Proposition. Let f : [ a, b ] ! R be of bounded variation. Then for x 2 [ a, b ], put V ( x ) = V f ( a, x ) (note that V ( a ) = 0), and let D ( x ) = V ( x ) - f ( x ). Then V and D are both increasing functions. Proof. Let a x < y b . We first prove that V f ( a, y ) = V f ( a, x ) + V f ( x, y ). Take a partition a = x 0 < x 1 < · · · < x n = x = y 0 < y 1 < · · · < y m = y . Then V f ( a, y ) n X k =1 | f ( x k ) - f ( x k - 1 ) | + m X j =1 | f ( y j ) - f ( y j - 1 ) | This is true for all partitions of [ a, x ] and [ x, y ], so V f ( a, y ) V f ( a, x ) + V f ( x, y ). Now take any partition P of [ a, y ]. Let P 0 be the refined partition that includes x . Notice that X P Δ f X P 0 Δ f V f ( a, x ) + V f ( x, y ) Since this is true for any partition P , V f ( a, y ) V f ( a, x ) + V f ( x, y ). Now consider V ( y ) - V ( x ) = V f ( a, y ) - V f ( a, x ) = V f ( x, y ) 0, and thus V ( y ) V ( x ) when y x . Also D ( y ) - D ( x ) = V ( y ) - f ( y ) - [ V ( x ) - f ( x )] = V f ( x, y ) - [ f ( y ) - f ( x )] | f ( y ) - f ( x ) | - [ f ( y ) - f ( x )] 0 so D ( y ) D ( x ) whenever y x . Corollary. f is of bounded variation if and only if f is a di erence of increasing functions. Proof. If f is of bounded variation, then f = V - D where V, D are increasing. Conversely if f = g - h where g and h are increasing functions, then g and h are of bounded variation, so f = g - h is of bounded variation. Corollary. If f is of bounded variation, then σ n f ( t 0 ) converges. Proof. Write f = V - D where V and D are increasing. V and D only have jump discontinuities, so f only has jump discontinuities. Thus f ( t + 0 ) and f ( t - 0 ) exist. Thus σ n f ( t 0 ) ! f ( t + 0 )+ f ( t - 0 ) 2 by Fejer’s theorem. Note. In fact, if f is continuous then V and D are continuous.
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