= Solution
Direct evaluation gives the first <Fourier transform>:
$$
\widetilde f(k)=\int_{-1/2}^{1/2}e^{-ikx}\,dx=\boxed{\frac{2\sin(k/2)}{k}},\qquad\boxed{\widetilde f(0)=1}.
$$
For the even <triangular function>, the odd sine part integrates to zero. Two elementary <integrations by parts> or one evaluation of the cosine integral give
$$
\widetilde g(k)=2\int_0^1(1-x)\cos(kx)\,dx=\frac{2(1-\cos k)}{k^2}=\boxed{\frac{4\sin^2(k/2)}{k^2}},\qquad\boxed{\widetilde g(0)=1}.
$$
The values at zero are the respective areas, agreeing with the removable limits. Therefore $\widetilde g=\widetilde f^{\,2}$. By the <convolution theorem> and uniqueness of the <Fourier transform>,
$$
\boxed{(f*f)(x)=g(x)=\begin{cases}1-|x|,&|x|<1,\\0,&|x|\geq1.\end{cases}}
$$
There is also a direct interpretation: the <convolution> is the length of overlap of the intervals $(-1/2,1/2)$ and $(x-1/2,x+1/2)$, giving exactly that triangular profile.
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