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Edited to clarify question and give accepted answer in context.

Looking at the table of Fourier Transforms on Wikibooks line 14 gives the Fourier transform of the triangle function $\left(1-\frac{2|t|}{\tau}\right)\mathrm{rect}\left(\frac{t}{\tau}\right)$ as $\frac{\tau}{2}\mathrm{sinc}^2\left(\frac{\tau\omega}{4\pi}\right)$.

The triangle function is the convolution of two rectangle functions $\mathrm{rect}\left(\frac{t}{\tau/2}\right)$.

Each rectangle function, by Line 12 has a Fourier transform $\frac{\tau}{2}\mathrm{sinc}\left(\frac{\tau\omega}{4\pi}\right)$.

The Fourier transform of the convolution should be the squared term $\frac{\tau^2}{4}\mathrm{sinc}^2\left(\frac{\tau\omega}{4\pi}\right)$ which is off by a factor of $\tau/2$.

Answer by @dfnu

The convolution of the two rectangle functions is the triangle function multiplied by $\tau/2$; which is the area of the product of the two rectangles for zero lag. To get the triangle function given, the convolution must be divided by $\tau/2$.

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I think you are missing factor $\frac2\tau$.

If you convolve $\operatorname{rect}\left(\frac{2t}\tau\right)$ with itself, you'll obtain a Fourier transform equal to $\frac{\tau^2}{4}\operatorname{sinc}^2\left(\frac{\tau\omega}{4\pi}\right)$.

However, this convolution does not give as a result the triangular function your mentioning. The convolution in $0$ is equal to the area of the rectangle, i.e. $\frac\tau 2$, whereas the triangle in $0$ is equal to $1$. So the triangle is actually \begin{eqnarray} h(t) &=&\left(1-\frac{2|t|}{\tau}\right)\mathrm{rect}\left(\frac{t}{\tau}\right)=\\ &=& \frac2{\tau}\cdot \operatorname{rect}\left(\frac{2t}\tau\right) *\operatorname{rect}\left(\frac{2t}\tau\right), \end{eqnarray}

and its Fourier transform is $$H(\omega) =\frac2\tau \frac{\tau^2}{4}\operatorname{sinc}^2\left(\frac{\tau\omega}{4\pi}\right).$$

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