Characteristic function of the sum of two Uniform independent RVs - convolution theorem does not seem to work! Why? I just came to find this annoying case I do not seem to able to figure out.
Let's assume $X_{1}$ and $X_{2}$ are $Uniform (-m_{j},m_{j})$, $j = 1,2$ and independent; since they are independent the characteristic function of their sum should be equal to the product of their characteristic functions.
In other words let's consider  $X = X_{1} + X_{2}$, then $\phi_{X}(t) = \phi_{X_{1}}(t) \phi_{X_{2}}(t)$.
The characteristic function is easily calculated: $\phi_{X_{j}}(t) = \frac{\sin(t\ m_{j})}{t\ m_{j}}$, right?
So it should also be $\phi_{X}(t) = \frac{\sin(t\ m_{1})}{t\ m_{1}} \frac{\sin(t\ m_{2})}{t \ m_{2}}$; however this is wrong!
If one calculates the convolution of the two PDFs gets $f_{X}(x) = \frac{1}{\Sigma}\left(1-\frac{|x|}{\Sigma}\right)$ (triangular distribution) where $\Sigma = m_{1} + m_{2}$.
Applying the definition of characteristic function to the PDF of $X$ one gets: $\phi_{X}(t) = 2 \frac{1\ - \cos\left(t \Sigma\right)}{\Sigma^2 t^2}$ (after many simplifications) which is the correct one!
What am I doing wrong when multiplying the two individual characteristic functions? I must be applying the convolution theorem (for Fourier transforms, or more accurately, for characteristic functions) wrongly, but I cannot pinpoint what hypotesis is violated... Can somebody help please?
 A: For $X\sim \textrm{Uniform}[-m,m]$ we have
$$E[e^{i\xi X}]=\frac{1}{2m}\int_{-m}^me^{i\xi x}dx=\frac{1}{2m}\frac{e^{i\xi m}-e^{-i\xi m}}{i\xi}=\frac{\sin(m\xi)}{m\xi}$$
So if $X_1,X_2$ are independent and uniform as above with parameters $m_1,m_2>0$:
$$E[e^{i\xi(X_1+X_2)}]=E[e^{i\xi X_1}]E[e^{i\xi X_2}]=\frac{\sin(m_1\xi)\sin(m_2\xi)}{\xi^2 m_1m_2}$$
Now note:
$$\mathbf{1}_{[-m_2,m_2]}(z-x)=\begin{cases}
1&-m_2\leq z-x\leq m_2\\
0&\textrm{otherwise}
\end{cases}=\begin{cases}
1&m_2+z\geq x\geq z-m_2\\
0&\textrm{otherwise}
\end{cases}$$
If we convolve the densities we get for $z \in [-m_1-m_2,m_1+m_2]$
$$\begin{aligned}f_{X_1+X_2}(z)&=\frac{1}{4m_1m_2}\int_\mathbb{R}\mathbf{1}_{[-m_1,m_1]}(x)\mathbf{1}_{[-m_2,m_2]}(z-x)dx=\\
&=\frac{1}{4m_1m_2}\int_\mathbb{R}\mathbf{1}_{[-m_1,m_1]}(x)\mathbf{1}_{[z-m_2,z+m_2]}(x)dx=\\
&=\frac{1}{4m_1m_2}(\min(z+m_2,m_1)-\max(-m_1,z-m_2))\end{aligned}$$
and $f_{X_1+X_2}(z)=0,\,\forall z \notin [-m_1-m_2,m_1+m_2]$. Example with $(m_1,m_2)=(2,3)$ with numerical Fourier transform of the pdf:

