(New page: Let <math>f\in L^1(\mathbb{R})</math>. Show that <math>\hat{f}(x)</math> is continuous and <math>\lim_{|x|\to\infty} \hat{f}(x)=0</math>. Proof: To show continuity, we only need to show t...) |
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Proof: To show continuity, we only need to show that if <math>x_k\to x</math> then <math>\hat{f}(x_k)\to\hat{f}(x)</math> | Proof: To show continuity, we only need to show that if <math>x_k\to x</math> then <math>\hat{f}(x_k)\to\hat{f}(x)</math> | ||
| − | <math>\lim_{k\to\infty}\hat{f}(x_k)=\lim_{k\to\infty}\int e^{- | + | <math>\lim_{k\to\infty}\hat{f}(x_k)=\lim_{k\to\infty}\int e^{-ix_kt}f(t)dt = \int e^{-ixt}f(t)dt = \hat{f}(x)</math> |
We can pass this limit through the integral since <math>\hat{f}</math> is dominated by <math>f\in L^1</math> | We can pass this limit through the integral since <math>\hat{f}</math> is dominated by <math>f\in L^1</math> | ||
Revision as of 15:44, 28 July 2009
Let $ f\in L^1(\mathbb{R}) $. Show that $ \hat{f}(x) $ is continuous and $ \lim_{|x|\to\infty} \hat{f}(x)=0 $.
Proof: To show continuity, we only need to show that if $ x_k\to x $ then $ \hat{f}(x_k)\to\hat{f}(x) $
$ \lim_{k\to\infty}\hat{f}(x_k)=\lim_{k\to\infty}\int e^{-ix_kt}f(t)dt = \int e^{-ixt}f(t)dt = \hat{f}(x) $
We can pass this limit through the integral since $ \hat{f} $ is dominated by $ f\in L^1 $
