Uniform boundedness principle theorem suppose x y are

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Uniform Boundedness Principle (Theorem). Suppose X, Y are Banach spaces. Let F be a family of bounded linear maps X ! Y . If for every x 2 X , sup F 2 F k F ( x ) k Y < 1 , then sup F 2 F k F k op < 1 . Proof. See PMATH 451. Corollary. Suppose X, Y are Banach spaces. Let F be a family of bounded linear maps X ! Y . If sup F 2 F k F k op = 1 , then sup F 2 F k F ( x ) k = 1 for some x 2 X .
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3 FOURIER ANALYSIS 39 Lecture 26: July 6 The uniform boundedness principle will be proved in PMATH 451. Today, we look at two applications of the UBP. Application 1. Consider X = Y = L 1 ( T ). Let F = { S N : L 1 ( T ) ! L 1 ( T ) , N 2 N } . Recall that S N ( F ) = P N n = - N ˆ f ( n ) e inx . This is a bounded linear map, since we saw previously that k S N k op 2 N + 1 < 1 for each N . We will show that k S N k op = k D N k L 1 c log N for some constant c > 0 that is independent of N . Hence sup S N 2 F k S N k op = sup N k S N k op sup N c log N = 1 . Thus by UBP there is a function f 2 L 1 ( T ) such that sup S N 2 F k S N ( f ) k L 1 = 1 . Thus sup N k P N - N ˆ f ( n ) e inx k 1 = 1 , so ( P N - N ˆ f ( n ) e inx ) 1 n =1 is divergent in L 1 . Thus there is a function f 2 L 1 whose Fourier series diverges in L 1 . We now show k D N k L 1 c log N . We use D N ( t ) = sin( N + 1 2 ) t sin 1 2 t . k D N k 1 = Z T | D N ( t ) | dm ( t ) = 1 2 Z 2 0 sin( N + 1 2 ) t sin t 2 dt Let y = t 2 so dy = 1 2 dt and k D N k 1 = 1 Z 0 sin(2 N + 1) y sin y dy 1 Z / 2 0 sin(2 N + 1) y sin y dy Note that for y 2 [0 , 2 ], we have 1 | y sin y | 2 , so 1 | sin y | 1 | y | . Thus k D N k 1 1 Z / 2 0 sin(2 N + 1) y | y | dy Let x = (2 N + 1) y , so dx = (2 N + 1) dy . Thus k D N k 1 1 Z (2 N +1) 2 0 | sin x | | x 2 N +1 | dx 2 N + 1 = 1 2 N X j =0 Z ( j +1) 2 j 2 | sin x | x dx 1 2 N X j =0 1 ( j + 1) 2 Z ( j +1) 2 j 2 | sin x | dx = 1 2 2 N X j =0 1 j + 1 Z / 2 0 sin x dx = 2 2 2 N X j =0 1 j + 1 = 2 2 2 N +1 X j =1 1 j 2 2 Z 2 N +1 1 1 t dt = 2 2 log(2 N + 1) c log N Application 2. We will now show that there is a continuous f such that S N f (0) diverges, so the Fourier series of a continuous does not necessarily converge pointwise. Step 1 Show 9 ( g n ) C ( T ) such that k g n k 1 1 and | S n ( g n )(0) | 1 2 k D n k 1 . Proof. S n ( g n )(0) = 1 2 R 2 0 g n ( t ) D N ( t ) dt . Note that D n ( t ) is real valued, and since it is a trig polynomial, it only changes sign finitely many times. We want g n ( t ) sgn( D n ( t )), but we need g n to be continuous. Define g n ( t ) = sgn( D n ( t )) except on little intervals around the zeros of D n ( t ), so that we can make g n continuous with | g n ( t ) | 1. Make the intervals so small that the sum of their lengths is at most " n ( " to be determined later).
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3 FOURIER ANALYSIS 40 Let J denote the union of these intervals. We then have | S n ( g n )(0) | = 1 2 Z 2 0 g n ( t ) D n ( t ) dt = 1 2 Z J g n ( t ) D n ( t ) dt + 1 2 Z [0 , 2 ] \ J g n ( t ) D n ( t ) dt = 1 2 Z J g n ( t ) D n ( t ) dt + 1 2 Z [0 , 2 ] \ J | D n ( t ) | dt = 1 2 Z 2 0 | D n ( t ) | dt - 1 2 Z J | D n ( t ) | dt + 1 2 Z J g n ( t ) D n ( t ) dt ≥ k D n k 1 - k D n k 1 1 2 Z J 1 dt - 1 2 Z J | g n || D n | = k D n k 1 - k D n k 1 ` ( J ) 2 - 1 2 k g n k 1 k D n k 1 Z J 1 dt ≥ k D n k 1 - (2 n + 1) " n 1 2 - 1 2 (2 n + 1) " n ≥ k D n k 1 - 2 " 1 2 k D
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