The one-dimensional endpoint estimate on a bounded interval

The fundamental-theorem identity and its immediate consequence. By [F2] the representative $u^{\ast }$ is absolutely continuous on the compact interval $[a,a+\epsilon ]\subseteq [a,b]$ and $u^{\ast }(t)-u^{\ast }(a)={\int }_a^t{u}^{\mathrm{\prime }}$ for every $t\in [a,a+\epsilon ]$, so $u^{\ast }(a)-u^{\ast }(t)=-{\int }_a^t{u}^{\mathrm{\prime }}$ and therefore $\mathrm{\mid }{u}^{\ast }(a)\mathrm{\mid }\le \mathrm{\mid }{u}^{\ast }(t)\mathrm{\mid }+{\int }_a^t\mathrm{\mid }{u}^{\mathrm{\prime }}\mathrm{\mid }\le \mathrm{\mid }{u}^{\ast }(t)\mathrm{\mid }+{\int }_a^{a+\epsilon }\mathrm{\mid }{u}^{\mathrm{\prime }}\mathrm{\mid }$ for every $t\in [a,a+\epsilon ]$.

The case $p=1$. Integrating the pointwise inequality of step 1.1 over $t\in (a,a+\epsilon )$ and dividing by $\epsilon >0$ gives $\mathrm{\mid }{u}^{\ast }(a)\mathrm{\mid }\le {\epsilon }^{-1}{\int }_a^{a+\epsilon }\mathrm{\mid }{u}^{\ast }\mathrm{\mid }+{\int }_a^{a+\epsilon }\mathrm{\mid }{u}^{\mathrm{\prime }}\mathrm{\mid }$, which is (ii) at the left endpoint.

The case $1<p<\mathrm{\infty }$. Raising the pointwise inequality of step 1.1 to the $p$-th power and using $\mathrm{\mid }x+y{\mathrm{\mid }}^p\le 2^{p-1}(\mathrm{\mid }x{\mathrm{\mid }}^p+\mathrm{\mid }y{\mathrm{\mid }}^p)$ gives $\mathrm{\mid }{u}^{\ast }(a){\mathrm{\mid }}^p\le 2^{p-1}(\mathrm{\mid }{u}^{\ast }(t){\mathrm{\mid }}^p+({\int }_a^{a+\epsilon }\mathrm{\mid }{u}^{\mathrm{\prime }}\mathrm{\mid }{)}^p)$ for every $t\in (a,a+\epsilon )$. Integrating in $t$ and dividing by $\epsilon$ yields $\mathrm{\mid }{u}^{\ast }(a){\mathrm{\mid }}^p\le 2^{p-1}({\epsilon }^{-1}{\int }_a^{a+\epsilon }\mathrm{\mid }{u}^{\ast }{\mathrm{\mid }}^p+({\int }_a^{a+\epsilon }\mathrm{\mid }{u}^{\mathrm{\prime }}\mathrm{\mid }{)}^p)$, and Holder's inequality [F4] converts the last term into $({\int }_a^{a+\epsilon }\mathrm{\mid }{u}^{\mathrm{\prime }}\mathrm{\mid }{)}^p\le {\epsilon }^{p-1}{\int }_a^{a+\epsilon }\mathrm{\mid }{u}^{\mathrm{\prime }}{\mathrm{\mid }}^p$. This is (i) at the left endpoint.

The right endpoint. Put $\tilde{u}(t):=u(a+b-t)$ for $t\in I$; then $t\mapsto u^{\ast }(a+b-t)$ is an absolutely continuous representative of $\tilde{u}$ whose classical derivative exists almost everywhere and equals $-u^{\mathrm{\prime }}(a+b-t)\in L^p(I)$ (the weak derivative class of $\tilde{u}$, by the classical chain rule and [F1]), so $\tilde{u}\in W^{1,p}(I;\mathbb{K})$ with weak derivative $-u^{\mathrm{\prime }}(a+b-\cdot )$, and its absolutely continuous representative takes the value $u^{\ast }(b)$ at $t=a$. Applying steps 2.1 and 2.2 to $\tilde{u}$ on $(a,a+\epsilon )$ gives the same inequalities with $u^{\ast }(b)$ on the left and the integrals over $(b-\epsilon ,b)$ on the right, because $\tilde{u}$ and its weak derivative on $(a,a+\epsilon )$ correspond to $u^{\ast }$ and $u^{\mathrm{\prime }}$ on $(b-\epsilon ,b)$ under the reflection.

Representative independence and the multiplied form. If $v$ is another representative of the class that is absolutely continuous on compact subintervals, then $v-u^{\ast }$ is constant by [F2]'s identity for both representatives with the same weak derivative class; since $v=u^{\ast }$ almost everywhere that constant is $0$ (the interval is nonempty), so the endpoint values of the absolutely continuous representative are determined by the class of $u$. Multiplying (i) and (ii) by $\epsilon$ gives the displayed $C_p$-form $\epsilon \mathrm{\mid }{u}^{\ast }(a){\mathrm{\mid }}^p\le 2^{p-1}({\int }_a^{a+\epsilon }\mathrm{\mid }{u}^{\ast }{\mathrm{\mid }}^p+{\epsilon }^p{\int }_a^{a+\epsilon }\mathrm{\mid }{u}^{\mathrm{\prime }}{\mathrm{\mid }}^p)$, which at $p=1$ reads $\epsilon \mathrm{\mid }{u}^{\ast }(a)\mathrm{\mid }\le {\int }_a^{a+\epsilon }\mathrm{\mid }{u}^{\ast }\mathrm{\mid }+\epsilon {\int }_a^{a+\epsilon }\mathrm{\mid }{u}^{\mathrm{\prime }}\mathrm{\mid }$.