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I am trying to answer the following:

Is $\sum_{n=1}^{\infty}\frac{\cos\left(nt\right)}{n^{3/2}}$ a Fourier series of a function in $C^{1}\left(\mathbb{T}\right)$ where $\mathbb T$ is the circle.

Attempt:

By the Weierstrass-$M$ test, it's easy to show that the series is uniformly convergent and moreover it's a series of continuous functions that converges uniformly. Therefore, it converges to a continuous function.

I know that uniform convergence implies convergence in $L^2(\mathbb T)$ thus if $\sum_{n\in\mathbb{Z}}c_{n}e^{int}\overset{\text{uniformly convergent}}{=}f\left(t\right)$ then $c_{n}=\hat{f}\left(n\right)$ by the uniqueness of the Fourier coefficients, where $\hat f(n)$ are the Fourier coefficients. I concluded that indeed the series is a Fourier series of a function in $C(\mathbb T)$.

About the differentiability, similarly to here: the derivatives series is not uniformly convergent, I managed to show that there is no uniform convergence for the series $S_{n}'=\left(\sum_{k=1}^{n}\frac{\cos\left(nt\right)}{n^{3/2}}\right)^{'}=\sum_{k=1}^{n}\frac{-\sin\left(nt\right)}{\sqrt{n}}$ thus it won't imply anything about the original series.

I was also trying to calculate the derivative using its definition, but failed.

This was from an exam in harmonic analysis. I am more interested in answers that solves it using tools from harmonic analysis rather than answers using stuff learned in calculus 2 such as calculating the derivative by definition if possible.

EDIT: I do think I managed to show its not differentiable at $0$ so some feedback would be helpful moreover again if there is some method to prove using harmonic analysis I am more interested in that.

Set $f(t)=\sum_{n=1}^{\infty}\frac{\cos\left(nt\right)}{n^{\frac{3}{2}}} $ thus: $ \left|\frac{f(th)-f(0)}{h}\right|=\left|\frac{\sum_{n=1}^{\infty}\frac{\cos\left(nh\right)}{n^{\frac{3}{2}}}-\sum_{n=1}^{\infty}\frac{1}{n^{\frac{3}{2}}}}{h}\right|=\left|\frac{1}{h}\cdot\sum_{n=1}^{\infty}\frac{\cos\left(nh\right)-1}{n^{\frac{3}{2}}}\right|=\left|\frac{1}{h}\right|\cdot\sum_{n=1}^{\infty}\frac{1-\cos\left(nh\right)}{n^{\frac{3}{2}}}$.

set $ h=\frac{\pi}{2^{s+1}}$ as $h$ approaches $0$ $s$ approaches $\infty$ then if: $2^{s}<n<2^{s+1}\implies\frac{\pi}{2}<nh<\pi\implies\cos\left(nh\right)\le0$ Hence:

$\left|\frac{1}{h}\right|\cdot\sum_{n=1}^{\infty}\frac{1-\cos\left(nh\right)}{n^{\frac{3}{2}}}\ge\left|\frac{1}{\frac{\pi}{2^{s+1}}}\right|\cdot\sum_{2^{s}<n<2^{s+1}}\frac{1-\cos\left(nh\right)}{n^{\frac{3}{2}}}\ge\frac{2^{s+1}}{\pi}\cdot\sum_{2^{s}<n<2^{s+1}}\frac{1}{n^{\frac{3}{2}}}$

$\ge\frac{2^{s+1}}{\pi}\cdot\sum_{2^{s}<n<2^{s+1}}\frac{1}{\left(2^{s+1}\right)^{\frac{3}{2}}}=\frac{1}{\pi}\cdot\sum_{2^{s}<n<2^{s+1}}\frac{1}{2^{\frac{3}{2}}}=\frac{1}{\pi}\cdot\left(2^{s+1}-2^{s}\right)\cdot\frac{1}{2^{\frac{3}{2}}}$

$=\frac{1}{\pi}\cdot\left(2^{s+1}-2^{s}\right)\cdot\frac{1}{2^{\frac{3}{2}}}=\frac{1}{\pi}\cdot2^{s}\left(2-1\right)\cdot\frac{1}{2^{\frac{3}{2}}}=\frac{1}{\pi}\cdot2^{s}\cdot\frac{1}{2^{\frac{3}{2}}}\overset{s\to\infty}{\longrightarrow}\infty$

Gardosh
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  • Perhaps $$f(t)=\sum\limits_{n=1}^{\infty} \frac{\cos(n t)}{n^{3/2}}=\frac{1}{2} \left(\text{Li}{\frac{3}{2}}\left(e^{-i t}\right)+\text{Li}{\frac{3}{2}}\left(e^{i t}\right)\right)$$ provides some insight where $\text{Li}_n(z)$ is the polylogarithm function. The function $f(t)$ is a $2 \pi$-periodic even function of $t$ with inflection points at $t=2 \pi n$ where $n\in\mathbb{Z}$. – Steven Clark Mar 16 '25 at 17:25
  • @StevenClark I don't think that I have the tools to prove that also I have no knowledge about that function and I should be able to solve the question without it... – Gardosh Mar 16 '25 at 17:29
  • Perhaps you could work at showing $$\underset{t\to 0^-}{\text{lim}}f'(t)=\infty$$ and $$\underset{t\to 0^+}{\text{lim}}f(t)=-\infty$$ where $$f'(t)=\sum_{n=1}^{\infty} -\frac{\sin(n t)}{\sqrt{n}}.$$ – Steven Clark Mar 16 '25 at 18:04
  • Note that $$f'(t)=\sum_{n=1}^{\infty} -\frac{\sin(n t)}{\sqrt{n}}=-\frac{1}{2} i \left(\sum_{n=1}^{\infty} \frac{e^{-i n t}}{\sqrt{n}}-\sum_{n=1}^{\infty} \frac{e^{i n t}}{\sqrt{n}}\right)=-\frac{1}{2} i \left(\text{Li}{\frac{1}{2}}\left(e^{-i t}\right)-\text{Li}{\frac{1}{2}}\left(e^{i t}\right)\right)$$ is consistent with the identity $$\sin(n t)=\frac{1}{2} i \left(e^{-i n t}-e^{i n t}\right)$$ and the definition $$\text{Li}{\alpha}(z)=\sum\limits{n=1}^{\infty} \frac{z^n}{n^{\alpha}}.$$ – Steven Clark Mar 16 '25 at 18:05
  • I have manged to show that u proposed but I don't understand how it follows from that that the series is not differentiable – Gardosh Mar 16 '25 at 18:07
  • Sorry, $f(t)$ should have been $f'(t)$ in $\underset{t\to 0^+}{\text{lim}}f(t)=-\infty$. The limits $\underset{t\to 0^-}{\text{lim}}f'(t)=\infty$ and $\underset{t\to 0^+}{\text{lim}}f'(t)=-\infty$ imply $f'(t)$ has a singularity at $t=0$. – Steven Clark Mar 16 '25 at 18:13
  • @StevenClark As I listened I have managed to show similar to here: https://math.stackexchange.com/questions/2912418/uniform-convergence-of-sum-n-1-infty-frac-sinnx-sqrtn-0-2-pi?noredirect=1&lq=1that the derivatives don't uniformly converge but that won't imply that the original series is not differentable, so I think I don't understand how what u have listed would help here – Gardosh Mar 16 '25 at 18:15
  • It seems to me you're focused on mode of convergence whereas I'm saying $f'(t)$ doesn't converge at $t=0$. – Steven Clark Mar 16 '25 at 18:20
  • Under the edit I have showed (or at least I think I did) that the dervative at 0 is infinte, what I don't understand is that you said: $f'(t)=\sum_{n=1}^{\infty} -\frac{\sin(n t)}{\sqrt{n}}.$ Clearly since I proved that the derivative don't uniformly converge I need to use by defintion of the derivative – Gardosh Mar 16 '25 at 18:23
  • $\underset{t\to 0^-}{\text{lim}}f'(t)=\infty$ (where $t$ approaches zero from below), but $\underset{t\to 0^+}{\text{lim}}f'(t)=-\infty$ (where $t$ approaches zero from above). – Steven Clark Mar 16 '25 at 18:27
  • I don't understand tho are u saying that I should differntable element by element but that won't tell me anything since there are sequences of differentiable functions that uniformly converge to a differentiable function, but the derivatives of which don't converge to the derivative of the limit function as shown here: https://math.stackexchange.com/questions/4669/example-function-sequence-uniformly-converges-its-derivatives-dont so I must use the defination of the dervative – Gardosh Mar 16 '25 at 18:29

1 Answers1

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Assume that $f(t)=\sum_{n=1}^\infty\frac{\cos(nt)}{n^{3/2}}$ was $C^1$. By integration by parts, we obtain $$\widehat{f'}(n)=\pm\frac{i}{2|n|^{1/2}}\qquad(n\neq 0).\tag{$*$} $$ But then Parseval's identity would imply $f'\notin L^2(\mathbb{T})$, which contradicts the assumption that $f'$ is continuous on $\mathbb{T}$.

(Edited) To prove $(*)$, note that $$\widehat{f'}(n)=\frac{1}{2\pi}\int_0^{2\pi}f'(t)e^{-int}dt=\frac{1}{2\pi}\left(f(2\pi)-f(0)+in\int_0^{2\pi}f(t)e^{-int}dt\right)=in\widehat{f}(n). $$

MintChocolate
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