Let $f(x)$ be the value of a trigonometric series, which converges uniformly on $\left[ -\pi, \pi\right]$. If I multiply $f(x)$ with $e^{iax}$ where $a\in\mathbb{N}$ will the result then be a trigonometric series which converges uniformly?
multiplication of a trigonometric series
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calculus
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0$a$ is fixed??? – 2012-06-11
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0a is a constant. – 2012-06-11
1 Answers
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Let $f_n$ denote the n-th partial sum of the series. We have $f_n \to f$ uniformly on $[-\pi,\pi].$ Clearly, $e^{iax} f_n(x) \to e^{iax} f(x)$ pointwise, and this is uniform since $$ \sup_{x\in [-\pi,\pi]} \| e^{iax}f_n(x) - e^{iax}f(x)\| =\sup_{x\in [-\pi,\pi]} \| f_n(x) - f(x)\| \to 0 \text{ as } n\to \infty.$$
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0This is good! This might be trivial, but I still need to know if the function $e^{iax}f(x)$ is the value of trigonometric series. – 2012-06-11
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0I'm not 100% sure what you are asking for clarification on. We showed the uniform convergence of $e^{iax} f_n(x) $ to $e^{iax} f(x)$, which implies pointwise convergence as well. More generally, if $f_n\to f$ and $g_n \to g$ pointwise, then $f_n g_n\to fg$ pointwise. Does this resolve your issue? – 2012-06-11
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0Given a trigonometric series, multiply it with $e^{iax}$ is the result then a trigonometric series? – 2012-06-11
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0@characters By the usual definition of a trigonometric series, yes ($a$ must be an integer). – 2012-06-11