= Solution
Insert part (i) in the <inverse Fourier transform>. The factor two in $\widehat T$ cancels half of the inverse normalization:
$$
\boxed{u(t,x)=\frac1\pi\int_{-\infty}^{\infty}
e^{-t\omega^2+ix\omega}\frac{\sin^2\omega}{\omega^2}\,d\omega.}
$$
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:
$$
u(t,x)=(q_t*T)(x)=\frac1{\sqrt{4\pi t}}\int_{-2}^2e^{-(x-y)^2/(4t)}T(y)\,dy.
$$
Since $q_t$ has integral one, write
$$
u(t,x)-T(x)=\int_{\mathbb R}q_t(y)\bigl(T(x-y)-T(x)\bigr)\,dy.
$$
The continuous compactly supported profile $T$ is <uniformly continuous>. Given $\epsilon>0$, choose $\delta>0$ so its change is below $\epsilon$ for $|y|<\delta$. That part of the integral is bounded by $\epsilon$. The other part is bounded by $2\|T\|_\infty\int_{|y|\geq\delta}q_t(y)\,dy$, 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 $t\downarrow0$ 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.
\Image[/past-exam-of-the-algebra-1-and-differential-equations-1-course-of-the-university-of-oxford/2014-triangular-heat-flow.png]
{title=The triangular initial profile smooths and spreads under heat flow}
{description=Original plots of the initial triangular profile and its convolution with the Gaussian heat kernel at three positive times. Each curve has total integral two on the real line.}
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