Smooth integral functionals approximate bounded functionals on compact sets

Empty $K$ or $L=0$ is handled by $\kappa =0$. Otherwise [F1] permits choosing $t>0$ with ${\sup }_K\mathrm{\parallel }{S}_{t}a-a{\mathrm{\parallel }}_X<\epsilon \mathrm{/}(2\mathrm{\parallel }L\mathrm{\parallel })$. For fixed $t$, the map $y\mapsto k_t(\cdot ,y){\mathrm{\mid }}_D$ is ambient $C^{\mathrm{\infty }}$ with values in $X$: every mixed $x$ and $y$ derivative is uniformly continuous on compact products, so the ordinary finite-dimensional Taylor remainders in $y$, uniformly for $x$ derivatives through order $m$, tend to zero in the $C^m$ norm and hence in the $X$ norm. Thus ${\kappa }_t(y)=L(k_t(\cdot ,y){\mathrm{\mid }}_D)$ is ambient smooth, with $D_y^{\alpha }{\kappa }_t=L(D_y^{\alpha }{k}_t(\cdot ,y){\mathrm{\mid }}_D)$. It is bounded on $\bar{D}$.

Compactness of $K$ in $X$ implies its compactness in $L^1(D)$ by [F2]. Given $\theta >0$, take a finite $\theta \mathrm{/}2$ net $a_1,\dots ,a_s$ in that $L^1$ image. By integrability and an exhaustion of $D$ by compact subsets, there is one compact $F\subset D$ with ${\int }_{D\setminus F}\mathrm{\mid }{a}_i\mathrm{\mid }<\theta \mathrm{/}2$ for all $i$. Therefore ${\sup }_{a\in K}{\int }_{D\setminus F}\mathrm{\mid }a\mathrm{\mid }<\theta$. Choose a smooth cutoff $0\le \zeta \le 1$, compactly supported in $D$, equal to one on $F$; use [F3] for the one-set open cover $\{D\}$ of $F$. Then ${\sup }_K{\int }_D(1-\zeta )\mathrm{\mid }a\mathrm{\mid }<\theta$. This uses a finite net and finitely many integrable tails, including at $p=1$; it requires no $L^{\mathrm{\infty }}$ approximation of a dual coefficient.

For each $a\in X$, one has $L(S_{t}a)={\int }_D{\kappa }_t(y)a(y)dy$. To justify the identity without assuming a vector-integration theorem, partition a box containing $D$ into cells of diameter tending to zero and replace $k_t(\cdot ,y)$ on each cell by $k_t(\cdot ,y_l)$. Uniform continuity of every $x$ derivative through order $m$ makes the error in $X$ at most $c_X\eta \mathrm{\parallel }a{\mathrm{\parallel }}_1$, where $\eta \to 0$ is the uniform kernel derivative error and $c_X$ converts a $C^m$ bound to an $X$ bound. Applying bounded $L$ to the finite sums is legitimate by linearity. Their scalar integral errors are at most $\mathrm{\parallel }L\mathrm{\parallel }{c}_X\eta \mathrm{\parallel }a{\mathrm{\parallel }}_1$ as well, so passing to the limit proves the identity. Consequently ${\sup }_K\mathrm{\mid }L(a)-{\int }_D{\kappa }_{t}a\mathrm{\mid }<\epsilon \mathrm{/}2$.

If $\mathrm{\parallel }{\kappa }_t{\mathrm{\parallel }}_{L^{\mathrm{\infty }}(D)}=0$, choose $\kappa =0$. Otherwise use step 1.2 with $\theta =\epsilon \mathrm{/}(2\mathrm{\parallel }{\kappa }_t{\mathrm{\parallel }}_{\mathrm{\infty }})$ and set $\kappa =\zeta {\kappa }_t\in C_c^{\mathrm{\infty }}(D)$. The cutoff error satisfies ${\sup }_K\mathrm{\mid }{\int }_D({\kappa }_t-\kappa )a\mathrm{\mid }\le \mathrm{\parallel }{\kappa }_t{\mathrm{\parallel }}_{\mathrm{\infty }}{\sup }_K{\int }_D(1-\zeta )\mathrm{\mid }a\mathrm{\mid }<\epsilon \mathrm{/}2$. Combining with step 2.1 proves the claim.