= Solution
Let $(\phi_j)$ be a smooth <orthonormal eigenbasis> of the <positive Laplace-Beltrami operator>, with <eigenvalues> $(\lambda_j)$. The <orthonormal eigenbasis> defines
$$
e^{-t\Delta}f=\sum_j e^{-t\lambda_j}\langle f,\phi_j\rangle\phi_j.
$$
The large-$j$ lower bound of part (c), valid also after the finitely many zero modes of a disconnected <compact manifold>, gives $\lambda_j\geq a j^{2/n}$ for sufficiently large $j$. Hence
$$
\sum_j e^{-t\lambda_j}<\infty\qquad(t>0),
$$
by comparison with $\int_0^\infty e^{-ta x^{2/n}}\,dx$. The operator is positive and diagonal with these summable coefficients, so it is a <trace-class operator> and its <operator trace> is their sum.
Its proposed <heat kernel> is
$$
K_t(x,y)=\sum_j e^{-t\lambda_j}\phi_j(x)\overline{\phi_j(y)}.
$$
We verify smoothness and the kernel action. The <elliptic regularity> estimate established in Question 4(d) gives, by induction,
$$
\|\phi_j\|_{H^{2r}}\leq C_r(1+\lambda_j)^r.
$$
The <Sobolev embedding theorem> therefore gives, for every derivative order $a$, constants $C_a$, $r_a$ with $\|\phi_j\|_{C^a}\leq C_a(1+\lambda_j)^{r_a}$. Every differentiated term of the kernel series is bounded by $C_{a,b}e^{-t\lambda_j}(1+\lambda_j)^{r_a+r_b}$. An exponential absorbs any fixed power: for $t\geq\tau>0$ this is at most $C_{a,b,\tau}e^{-\tau\lambda_j/2}$, a summable bound. Time derivatives merely introduce more powers of $\lambda_j$. Thus the series converges uniformly with all spatial and time derivatives on $[\tau,\infty)\times M\times M$. In particular $K_t$ is smooth for every $t>0$.
Integrating its uniformly convergent series against $f\in L^2(M)$ is legitimate since the <compact manifold> has finite volume and $L^2(M)\subseteq L^1(M)$. It gives exactly the spectral expression above, so
$$
(e^{-t\Delta}f)(x)=\int_MK_t(x,y)f(y)\,d\mathrm{vol}_g(y).
$$
Finally $K_t(x,x)=\sum_j e^{-t\lambda_j}|\phi_j(x)|^2$ is nonnegative. Integrating term by term, either by uniform convergence or by <Tonelli theorem>, and using $\|\phi_j\|_2=1$ proves the <heat kernel trace formula>
$$
\boxed{\operatorname{Tr}e^{-t\Delta}=\sum_j e^{-t\lambda_j}=\int_MK_t(x,x)\,d\mathrm{vol}_g(x),\qquad t>0.}
$$
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