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I just started learning about the Fourier series, is this statement true or false?

Looking at $\mathcal {R}(-\pi,\pi).$ If $\hat{f}(k)=0$ for all $k <0$, then $f(x)\geq0$ for all $x$.

Philip730
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  • Are you requiring that $f(x)$ is real? – Zim Jun 16 '20 at 16:15
  • Does it make a difference? Could you answer it for both maybe? – Philip730 Jun 16 '20 at 16:35
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    If $f$ real, $\hat{f}(k)=\bar{ \hat{f}(-k)}$ so the OP condition implies $f$ real constant though it may be negative (note that the OP condition is unchanged under multiplication by a constant so the original question doesn;t make sense as given, but it should be interepreted as $f$ has constant sign; if $f$ is general, the condition just implies that $f \in H^1$ and there is no reason for either $ \Re f, \Im f$ to be of constant sign – Conrad Jun 16 '20 at 16:47
  • @Philip730 Unfortunately, since there isn't a well-defined order on the complex numbers, it's not entirely clear what you're asking if $f$ takes complex values. – Zim Jun 16 '20 at 17:52

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Notice that $\widehat{f}(k)=\frac{1}{2\pi}\int^\pi_{-\pi}e^{-iky}f(y)\,dy=\overline{\widehat{f(-k)}}$

So if $f$ is real, your condition implies that $\widehat{f}(k)=0$ for all $k$ and so $f=f(0)=\widehat{f}(0)$ a.s. So, unless $f(0)\geq0$, the answer to your question is general is NO.

Mittens
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