330 HW 04 - MAT 330: Fourier Series and Partial...

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MAT 330: Fourier Series and Partial Differential Equations Professor Sergiu Klainerman Problem Set 4 Rik Sengupta rsengupt@princeton.edu March 1, 2010 1. Stein and Shakarchi, p. 124, problem 11 Show that if u ( x,t ) = ( f * H t )( x ) where H t is the heat kernel, and f is Riemann integrable, then Z 1 0 | u ( x,t ) - f ( x ) | 2 dx 0 as t 0 . Solution. We have Z 1 0 | u ( x,t ) - f ( x ) | 2 dx = X n = -∞ | ˆ f ( n ) ˆ H t ( n ) - ˆ f ( n ) | 2 = X n = -∞ | ˆ f ( n ) 2 | ± ± ˆ H t ( n ) - 1 ± ± 2 = X | n |≥ N | ˆ f ( n ) 2 | ± ± ˆ H t ( n ) - 1 ± ± 2 + X | n | <N | ˆ f ( n ) 2 | ± ± ˆ H t ( n ) - 1 ± ± 2 , for some N N , N > 0. Fix ± > 0. Now, since f is integrable, we know, for large enough N , we can make | n |≥ N | ˆ f ( n ) 2 | arbitrarily small (an immediate consequence of Parseval’s identity). Furthermore, for this N , we have | ˆ H t ( n ) - 1 | 2 = | e - 4 π 2 n 2 t - 1 | → 0 as t 0, so in particular, the | ˆ H t ( n ) - 1 | term is always bounded by 1. Hence, for the first term above, we get X | n |≥ N | ˆ f ( n ) 2 | ± ± ˆ H t ( n ) - 1 ± ± 2 X | n |≥ N | ˆ f ( n ) 2 | · 1 = X | n |≥ N | ˆ f ( n ) 2 | < ± 2 . 1
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For the second term, we have a finite number of terms. Let M = max {| ˆ f ( n ) 2 |} for this finite number of terms. Furthermore, note that we are taking the limit as t goes to zero, and we have already shown above that the term | ˆ H t ( n ) - 1 | 2 = | e - 4 π 2 n 2 t - 1 | → 0 as t 0. So in particular, we can choose some t small enough so that | ˆ H t ( n ) - 1 | 2 = | e - 4 π 2 n 2 t - 1 | < ± 2(2 N - 1) M for all n . Then, we get X | n | <N | ˆ f ( n ) 2 | ± ± ˆ H t ( n ) - 1 ± ± 2 X | n | <N M ± ± ˆ H t ( n ) - 1 ± ± 2 < 2(2 N - 1) M X | n | <N 1 = 2(2 N - 1) M ·
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330 HW 04 - MAT 330: Fourier Series and Partial...

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