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I am unable to confirm the following. If a signal is odd, then does the following hold? $$a_k = -a_{-k}$$ If a signal is even, then: $$a_k = a_{-k}$$ If the aforementioned qualities hold, then is the following true? If a signal is odd and the period is $2k$, then $a_k = -a_{-k} = 0$.

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Your first two affirmations hold.

Let $f$ be $p$-periodic and $f \in L^1[-p/2,p/2]$. Then the Fourier coefficients of $f$ are $$ \hat{f}(k) = \frac{1}{p} \int_{-p/2}^{p/2} f(t) e^{-i2\pi k t/p} \,dt $$ If $f$ is odd then the change of variable $u=-t$ gives (we can keep the interval of integration $[-p/2,p/2]$ by $p$-periodicity) \begin{align} \hat{f}(k) &= -\frac{1}{p} \int_{-p/2}^{p/2} (-f(-t)) e^{i2\pi k t/p} \,dt \\ &= -\frac{1}{p} \int_{-p/2}^{p/2} f(t) e^{-i2\pi (-k) t/p} \,dt \\ &= -\hat{f}(-k) \end{align}

If $f$ is even then the same change of variable gives \begin{align} \hat{f}(k) &= \frac{1}{p} \int_{-p/2}^{p/2} f(-t) e^{i2\pi k t/p} \,dt \\ &= \frac{1}{p} \int_{-p/2}^{p/2} f(t) e^{-i2\pi (-k) t/p} \,dt \\ &= \hat{f}(-k) \end{align}

I don't think your third affirmation holds.

Take $f(t):=\sin(\pi t)$. Then $f$ is odd and of period $2\cdot1$. Yet, we can calculate $\hat{f}(1)=-i/2 \neq 0$.

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    But logically speaking, if the period is 2k, then, the value at -1 and 1 should be the same. But them being odd requires that the signs of the values are opposite. Hence, the only value that satisfies both is 0, right? Or am I erring in my logic? – 2017-02-16
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    @Christian Hmmm yes I think you are correct, my example is not an odd signal. Let me rethink this. – 2017-02-16
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    @Christian Do you agree with my new counter-example? – 2017-02-16
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    Oh, I see. Just to confirm, the notation with caret sign over the function signifies the Fourier coefficient at k=1 right? Not the actual function itself? – 2017-02-16
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    Also, can you please point out what I am missing in my logic? Because, I do agree with your counter-example. But that counter example implies that I'm missing some key information in my proof. – 2017-02-16
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    @Christian That is correct. $$\hat{f}(1) = \frac{1}{p} \int_{-p/2}^{p/2} f(t) e^{-i2\pi t/p} \,dt$$ which in my example reduces to $$\frac{1}{2} \int_{-1}^{1} \sin(\pi t) e^{-i\pi t} \,dt$$ [Wolfram alpha](http://www.wolframalpha.com/input/?i=1%2F2+*+int(sin(pi+t)+e%5E(-i+pi+t),+t%3D-1..1)) calculated it to be $-i/2$. – 2017-02-16
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    @Christian What proof are you talking about? What you said in your first comment, if I understood, is that if $f$ is odd and of period $2k$ then $f(k)=f(-k)=0$. I agree with that. By periodicity we have $f(k)=f(-k)$ and since $f$ is odd we have $f(k)=-f(-k)$. Putting those two conditions together gives $f(k)=-f(k)$, that is, $f(k)=0$ (and so $f(-k)=f(k)=0$). **Those are the values of the signal $f$ at $\pm k$, not of its Fourier coefficients $\hat{f}(\pm k)$** – 2017-02-16
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    So I guess my confusion is why is it that the arguments you just mentioned hold true for $f(k)$ but not for it's coefficients? Because can't the same arguments be made for the coefficients $a_k$ be made as well? – 2017-02-16
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    @Christian The Fourier coefficients will satisfy the odd property $\hat{f}(k)=-\hat{f}(-k)$, as I showed. However they won't be periodic in general. That's why it doesn't work. – 2017-02-16
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    Oh, I see. Thank you! That clears it up!! – 2017-02-16