smooth functions on compact Lie groups Let $G$ be a compact matrix (Lie) group. If $$f:G\longrightarrow \mathbb{C}$$ is a smooth function I would like to know if there are finite number of smooth functions $$f_{k}, \hat{f}_{k}:G\longrightarrow \mathbb{C} $$  such that $$f(g_1g_2)=\sum_k f_k(g_1)\hat{f}_{k}(g_2) $$ for all $g_1$, $g_2$ $\in$ $G$
 A: Here is a proof that there exists some $f:\mathbb S^1\to\mathbb R$ that does not admit such a product decomposition. What I will show is that the set of such functions is dense in the $\mathcal C^0$ topology, but it is not a constructive proof. As Qiaochu Yuan suggests, I think the exponential should be a counterexample when we see $\mathbb S^1$ as a subgroup of $\mathbb C^*$, but I don't know how to prove it. In the following, I will see $\mathbb S^1$ as $\mathbb R/\mathbb Z$.
I base my approach on the following property, that is forced by the formula you describe. Let $f:\mathbb S^1\to\mathbb R$ be a function such that
$$ f(x+y) = \sum_{k=1}^Kg_k(x)h_k(y). $$
If we fix $(x_1,\ldots,x_L)\in(\mathbb S^1)^L$, then for any $y\in\mathbb S^1$ the vector $(f(x_1+y),\ldots,f(x_L+y))\in\mathbb R^L$ belong to the $K$-dimensional subspace generated by the $(g_k(x_1),\ldots,g_k(x_L))$. Of course, if $L$ is a lot larger than $K$, then this should be highly unlikely.
Let us work on this fundamental property. Let $(B,|\cdot|_B)$ be a Banach space contained in $\mathcal C^\infty(\mathbb S^1,\mathbb R)$ such that $(B,|\cdot|_B)\to(\mathcal C^0,|\cdot|_\infty)$ is continuous dense.¹ Let $U_K$ be the set of $f\in B$ such that there exists $L\geq0$, $(x_1,\ldots,x_L)\in(\mathbb S^1)^L$ and $y_1,\ldots,y_K\in\mathbb S^1$ such that all the $x_\ell+y_k$ are distincts, and the $K$ vectors $(f(x_1+y_k),\ldots,f(x_L+y_k))$ span a subspace of $\mathbb R^L$ of dimension $K$ (in particular, $L\geq K$). Then the remark above says that if $f$ belongs to $U_K$, then it cannot be decomposed as a sum of at most $K-1$ products.

Lemma.
For all $K\geq0$, $U_K$ is open dense.

In particular, according to the Baire category theorem, the intersection of all the $U_K$ is dense, and no function in this intersection can satisfy a product decomposition formula as above.
The fact that $U_K$ is open comes from the fact that freedom is an open condition, and that the injection $B\to\mathcal C^0$ is continuous. The density is a bit more subtle. First, choose $(x_1,\ldots,x_K)$ and $(y_1,\ldots,y_K)$ as you please, provided the $x_\ell+y_k$ are all disjoint (note that $K=L$). Then the condition that $(f(x_1+y_k),\ldots,f(x_L+y_k))_k$ are independent amounts to saying that the determinant of the matrix
$$ (F_{k,\ell})_{0\leq k,\ell\leq K}:=\big(f(x_\ell+y_k)\big)_{0\leq k,\ell\leq K} $$
does not vanish.
Now the determinant of $F+t\vec F$, for some matrix $\vec F$ and $t\geq0$, is a polynomial in $t$, hence with respect to $t$ it can either be constant equal to zero or nonzero for all $t>0$ small enough. Of course there exists $\vec F$ such that $F+t\vec F$ is not constant equal to zero: for instance, fix $\vec F$ such that $F+\vec F$ is the identity. In particular, for such $\vec F$, $t\mapsto\det(F+tM)$ is nonzero for $t>0$ small enough. Let us fix such a matrix.
Choose $\vec f$ in $B$ such that $\vec f(x_\ell+y_k)$ is very close to $\vec F_{k,\ell}$ for all $(k,\ell)$, close enough that
$$ t\mapsto\det\big((f+t\vec f)(x_\ell+y_k)\big) $$
is not constant equal to zero (is it possible because $B$ is dense in $\mathcal C^0$). It is still a polynomial in $t$, so it is still nonzero for all $t>0$ small enough; it means that $f+t\vec f$ is in $U_K$ for all $t>0$ small enough. Of course $f+t\vec f$ converges to $f$ as $t$ goes to zero, so f is in the closure of $U_K$, and $U_K$ is indeed dense in $B$.

¹ For instance, $B$ is the closure of the trigonometric polynomials for the Hilbert norm
$$ |f|_B^2 = |a_0(f)|^2 + \sum_{k>0}2^k\left(|a_k(f)|^2+|b_k(f)|^2\right). $$
A: Here is an explicit counterexample, and indeed it is the exponential from $\mathbb U=\{z\in\mathbb C,|z|=1\}$ to $\mathbb C$. The ideas are similar to that of my other answer, but I feel like the result is different enough to be written independently.
Suppose the exponential satisfies such a product formula. Then it is easy to see that the function
$$ f:t\in\mathbb R\mapsto\exp(e^{it})\in\mathbb C $$
satisfies the same kind of relation:
$$ f(x+y) = \sum_{k=1}^Kg_k(x)h_k(y). $$
Take the $n$th derivative with respect to $x$, then the $m$th derivative with respect to $y$, and finally plug in the values $x=y=0$. We get
$$ f^{(n+m)}(0) = \sum_{k=1}^Kg_k^{(n)}(x)h_k^{(m)}(y). $$
In particular, for any fixed $m$, the vector $\big(f^{(m)}(0),\ldots,f^{(m+n)}(0)\big)$ belongs to the vector space generated by the vectors $\big(g_k^{(0)}(0),\ldots,g_k^{(n)}(0)\big)$ for $1\leq k\leq K$. Since this space must be of dimension at most $K$, it means that the $K+1$ vectors $\big(f^{(k)}(0),\ldots,f^{(k+K)}(0)\big)$, for $0\leq k\leq K$, are not free, hence the determinant of the $(K+1)\times(K+1)$ matrix
$$ M_K:=\big(f^{(n+m)}(0)\big)_{nm} $$
must vanish.
For now, we have not used anything about $f$ (apart from smoothness at zero). We will show that the determinant of $M_K$ is never zero for any $K$, which will conclude. This is where I make heavy use of Wikipedia.
It is not difficult to show by induction that the $n$th derivative of $f$ must be of the form $t\mapsto i^n\exp(e^{it})T_n(e^{it})$ for some polynomial $T_n$. It turns out that these polynomial are the so-called Touchard polynomials (for instance, we can show by induction that they satisfy the characterizing relations $T_{n+1}(X)=X\cdot(T_n(X)+T_n'(X))$, $T_0(X)=1$). The value of $f^{(n)}$ at zero is then $i^ne$ times the value of $T_n$ at one, so the determinant of $M_K$ is, up to a factor $(i^ne)^{K+1}$, the determinant of the matrix $A_K:=(T_{n+m}(1))_{nm}$.
The value at one of $T_n$ turns out to be the $n$th Bell number: $T_n(1)=B_n$, so $A_K$ is the matrix $(B_{n+m})_{nm}$ (it is not too difficult to see once we realise that the coefficients of the $T_n$ are the Stirling numbers of the second kind, and what they mean in combinatorial terms). One first way to prove it cannot be singular is to do a bit of research and see that it is precisely the matrix $\tilde A$ studied by M. Aigner in A Characterization of the Bell numbers,* who shows that
$$ \det A_K = \prod_{k=0}^Kk! $$
In particular, this determinant is never zero, providing $\exp:\mathbb U\to\mathbb C$ from satisfying a product formula like the one you describe.
In the comments below, Qiaochu Yuan suggested a very surprising second approach (at least to me), using probability. The first step is to realise that $B_n$ is the $n$th moment of a Poisson variable $X$ of parameter 1. This means that $A_K$ is the covariance matrix of $Y_K:=(X^0,\ldots,X^K)$, i.e. $A_K=\mathbb E[Y_KY_K^*]$. This matrix is real symmetric, so it will be non-singular provided the associated quadratic form is non-degenerate. In fact we can prove that $A_K$ is positive-definite; in other words, for any nonzero $p=(p_0,\ldots,p_K)$, we have $p^*A_Kp>0$. Indeed, in our case,
$$ p^*A_kp = \mathbb E[(p^*Y_K)(p^*Y_K)^*] = \mathbb E[|p^*Y_K|^2] = \mathbb E[|P(X)|^2], $$
where $P$ is the polynomial $P(x)=p_0+\cdots+p_Kx^K$. This expression is zero if and only if $P(X)=0$ almost surely, which means that either $X$ takes (almost surely) only a finite number of values or $P$ is constant equal to zero. Both possibilities are absurd, so $A_K$ is positive-definite, hence its determinant does not vanish. As above, this concludes (negatively) about the existence of a product formula for the exponential.
* At the time of writing, the article is freely available at this address.
