This question seem a bit elementary, but I find it more subtle than its looks. So, I post the question here.


Let $f(x,t) : [0,2\pi] \times [0,1] \to \mathbb{C}$ be a function such that $f(0,t)=f(2\pi,t)$ for all $t \in [0,1]$. Also, further suppose that $f(x,t)$ is $C^\infty$ in $x$ for each $t \in [0,1]$ and jointly continuous in $(x,t) \in [0,2\pi] \times [0,1]$.


Then, by Fourier expansion, I see that
\begin{equation}
f(x,t)=\sum\_{n \in \mathbb{Z}} a\_n(t) e^{in x}
\end{equation}
where each $a\_n(t)= \int\_0^{2\pi} f(x,t) e^{-inx} dx$ is continuous on $[0,1]$.


Now, by $C^\infty$ assumption with respect to $x$, we observe that $a\_n(t)=O\_t(1/\lvert n\rvert^k)$ for any $k \in \mathbb{N}$. Here I used the notation $O\_t$ to emphasize that this estimate depends on $t \in [0,1]$.


I wonder if $x$-derivatives of $f(x,t)$ are continuous in $t$ by any chance, since the Fourier coefficients $a\_n(t)$ are all continuous on the compact interval $[0,1]$.


It seems that
\begin{equation}
\lvert a\_n(t) \rvert \leq \frac{C\_{n,k}}{ \lvert n \rvert^k}
\end{equation}
for all $t \in [0,1]$, $n \in \mathbb{Z}$ and $k \in \mathbb{N}$ for some positive constants $C\_{n,k}$, but I do not see how to control $C\_{n,k}$'s.


Could anyone please help me?



If $f(x,t)=\sum_{n \in \mathbb{Z}} a_n(t) e^{in x}$ is $C^\infty$ in $x$ and all $a_n(t)$ continuous, $x$ derivatives of $f$ are continuous in $t$?
I am going to intepret the question as: given an $f$ satisfying the conditions in the question, must $\partial\_x f$ be continuous as a function on $[0,2\pi]\times [0,1]$?


@Echo's suggestion seems to give a concise, explicit counterexample. However, I thought it might be worth spelling out a line of thought that can be adapted to other contexts.




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The first thing to note is that Fourier analysis is really a red herring. Indeed, while we know that functions with a certain degree of smoothness have Fourier coefficients decaying at a certain rate, in general one loses information and hence cannot use the decay of Fourier coefficients to get back the same level of smoothness. (For instance $o(1/n)$ decay is satisfied by Fourier coefficients of all $C^1$ functions on the circle, but you cannot use $o(1/n)$ decay to deduce that the original function is in $C^1$.) Also, periodicity is not really important, since you are interested in continuity and smoothness which are local not global properties.


So I will produce an example of a function $F: [0,1]\times [0,1]\to {\mathbb R}$ which is (jointly) continuous, is $C^\infty$ in the first variable, but such that $\partial\_1 F$ is not jointly continuous on $[0,1]\times [0,1]$. Once you have such an $F$, one can extend it by reflection to get a function $F:[-1,1]\times [0,1] \to {\mathbb R}$ with all the same properties and also satisfying $F(-1,y)=F(1,y)$ for all $y$. It should then be clear how to rescale the domain to get a function that satisfies your original requirements.


The key is that once one starts trying to disprove the claim/question that you describe, it is natural to ask what a counterexample should look like. We want $\partial\_1 F$ to be discontinuous as a function on $[0,1]\times [0,1]$ even though $\partial\_1 F(\cdot, y)$ should be very nice for each fixed $y$. This suggests looking at one of the standard examples which shows separate continuity does not imply joint continuity. Let's pick the following function off the shelf:
$$
g(x,y) = \begin{cases} \dfrac{xy}{x^2+y^2} & \text{ if $(x,y) \neq (0,0)$ } \\ 0 & \text{ if $x=y=0$} \end{cases}
$$
This is not continuous at $(0,0)$, since $g(x,x)= \frac{1}{2}$ for all $x>0$.


Of course $g(\cdot ,0)$ is the zero function on $[0,1]$, while for any $y\in (0,1]$ the function $g(\cdot, y)$ is the quotient of a polynomial by a non-vanishing polynomial hence is smooth.


We now take this as our Ansatz for $\partial\_1 F$. That is, let us define
$$
F(x,y) = \int\_0^x g(u,y) \, du.
$$
Integration improves continuity/smoothness properties, so for each $y\in [0,1]$ we still have $F(\cdot y) \in C^\infty[0,1]$. Therefore all I need to do is to show the following result.


CLAIM. $F:[0,1]\times [0,1] \to {\mathbb R}$ is continuous.


Proof. Observe that $g$ is continuous at every point of $[0,1]\times [0,1]$ except the origin, and so the same behaviour is true for $F$, again because integrating will only improve things. It remains to show that $F$ is continuous at the origin.


Note that $F(0,\cdot)=0$. So $F(x,y) - F(0,0) = F(x,y) - F(0,y)$.
But since $0\leq g \leq 1$ on our given domain, this is bounded in modulus by $|x|$, and so as $(x,y)\to (0,0)$ we see that $|F(x,y)-F(0,0)| \leq |x| \to 0$, as required.
$\qquad\Box$




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As a final remark: I defined $F$ in this way to show that it is not necessary to have an "explicit formula" to do calculations and prove properties of $F$. However, of course we know by calculus how to "antidifferentiate" $g(\cdot, y)$, so we could take $F(x,y) = \frac{1}{2}y\log (x^2+y^2)- y\log y$.


