 Solution

Insert part (i) in the inverse Fourier transform. The factor two in $\hat{T}$ cancels half of the inverse normalization:The integrand has a removable singularity at zero. For $t>0$, the Gaussian controls every polynomial factor at infinity, so differentiation under the integral is valid. Differentiation in $t$ and twice in $x$ both multiply the integrand by $-{\omega }^2$, proving $u_t=u_{xx}$.

There is also a direct physical-space representation:Since $q_t$ has integral one, writeThe continuous compactly supported profile $T$ is uniformly continuous. Given $ϵ>0$, choose $\delta >0$ so its change is below $ϵ$ for $|y|<\delta$. That part of the integral is bounded by $ϵ$. The other part is bounded by $2\Vert T{\Vert }_{\infty }{\int }_{|y|\ge \delta }{q}_t(y)dy$, which tends to zero by the substitution $y=2\sqrt{t}z$ and the vanishing tails of the Gaussian integral. This proves $u(t,x)\to T(x)$ uniformly as $t\downarrow 0$ and verifies the initial condition.

Thus the formula supplies the full heat evolution of a triangular profile, including the corner points of the initial profile. Positive time immediately smooths those corners.