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On the real line, let $T(x)=\max (1-|x|/2,0)$. The Fourier transform of a triangular function gives $\hat{T}(\omega )=2{\sin }^2\omega /{\omega }^2$, with value two at zero. The Gaussian heat kernel has transform $e^{-t{\omega }^2}$. Hence $u=q_t\ast T$ solves the heat equation with initial profile $T$, and the inverse Fourier transform gives the displayed integral. Differentiating for positive time is justified by Gaussian decay. The uniform continuity of $T$ and the vanishing Gaussian mass outside any fixed neighborhood of zero prove uniform recovery of $T$ as $t\downarrow 0$. Convolution preserves total integral two, keeps the solution nonnegative, and smooths the corners for every positive time. Unlike the finite-interval problem in heat flow from a triangular initial temperature, this evolution has no endpoints through which heat can escape.

The profile is an even function, so its Fourier transform is real andFor $\omega \ne 0$, integration by parts givesThereforeAt zero, direct integration gives the area of the triangle, namely two; the displayed quotient has the same limit. In particular $\hat{T}(\omega )\ge 0$ for every real frequency. This is the Fourier transform of a triangular function.

One can also see the nonnegative square structurally: $T=\frac{1}{2}{1}_{[-1,1]}\ast 1_{[-1,1]}$, so the convolution theorem gives $\hat{T}=\frac{1}{2}(2\sin \omega /\omega {)}^2$.

PutThis function is bounded near zero and is $O(u^{-2})$ at infinity, so $H\in L^1(\mathbb{R})$ and the defining integral for $g$ converges absolutely.

For the triangular functiondirect integration givesThe Fourier inversion theorem, or equivalently the Fourier transform of a triangular function, therefore yieldsIn particular $g(x)=0$ whenever $|x|>1$. Finally,This also agrees with the Plancherel theorem applied to $h$ and $H$.

The even extension is the triangular function $f(x)=(1-|x|)$ for $|x|<1$ and zero outside. Its Fourier transform is real and even:At zero the removable limiting value is one, also obtained directly by integrating the triangle. This is the Fourier transform of a triangular function.

The triangular function is real and even, so its Fourier transform reduces to a cosine integral:For $k\ne 0$, one integration by parts gives ${\int }_0^1(1-x)\cos (kx)dx=(1-\cos k)/k^2$. Therefore the Fourier transform of a triangular function isThe zero-frequency value is the removable limit and also the area of the triangle. The transform is a squared sinc function, which explains its nonnegativity despite the oscillating Fourier exponential.

Evenness means the negative half of the function is determined as well: this is the triangular function $q(x)=1-|x|$ on $|x|\le 1$, zero outside. Its transform is real and even:Integration by parts, or integrating the two terms separately, givesThe apparent singularity at zero is removable, and its limiting value is the area under $q$. This is the Fourier transform of a triangular function.