Absolutely continuous representative of an L2 function with L2 weak derivative

[Reduction to the published corollary] The interval $(a,b)$ has finite Lebesgue measure $b-a$, and $1\le 2<\mathrm{\infty }$, so [F3] gives $u\in L^2(a,b)\subseteq L^1(a,b)$ and $u^{\mathrm{\prime }}\in L^2(a,b)\subseteq L^1(a,b)$, while $u^{\mathrm{\prime }}$ is the $L^2$ weak derivative of $u$ by [F1]. By [F4] the class $u$ lies in $W^{1,2}(a,b;\mathbb{C})$ with weak derivative $u^{\mathrm{\prime }}$.

[Existence and uniqueness of the representative] Apply [F2] with $I=(a,b)$ and $p=2$: there is exactly one representative $u^{\ast }$ of the class $u$ that is continuous, locally absolutely continuous and satisfies the integral formula. Since $I$ has finite endpoints, [F2] gives $u^{\mathrm{\prime }}\in L^1(a,b)$, the unique extension of $u^{\ast }$ to an absolutely continuous function on $[a,b]$, and the formula for all $x,y\in [a,b]$; the assignment $u\mapsto u^{\ast }$ is linear because the weak derivative and the integral are linear (One-dimensional Sobolev spaces on intervals records additivity of the weak derivative). This proves (a).

[Holder modulus] Let $x,y\in [a,b]$ with $x\le y$. If $x=y$, both sides are zero. If $x<y$, apply the complex integral triangle inequality in [F3] and then real Holder to $\mathrm{\mid }{u}^{\mathrm{\prime }}\mathrm{\mid }$ and $1$ on $(x,y)$, whose measure is $y-x$, to obtain $\mathrm{\mid }{u}^{\ast }(x)-u^{\ast }(y)\mathrm{\mid }=\mid {\int }_x^y{u}^{\mathrm{\prime }}\mid \le {\int }_x^y\mathrm{\mid }{u}^{\mathrm{\prime }}\mathrm{\mid }\le \mathrm{\parallel }{u}^{\mathrm{\prime }}{\mathrm{\parallel }}_{L^2(x,y)}\sqrt{y-x}\le \mathrm{\parallel }{u}^{\mathrm{\prime }}{\mathrm{\parallel }}_{L^2(a,b)}\sqrt{y-x};$ the case $x>y$ follows by exchanging $x$ and $y$. This is (b).

[Second-order representative] If additionally $u\in H^2(a,b)$, then $u^{\mathrm{\prime }}\in H^1(a,b)$ with weak derivative $D^{2}u\in L^2(a,b)$ by [F1]. Applying the already proved part (a) with $u^{\mathrm{\prime }}$ in place of $u$ supplies a continuous representative $w$ of the class $u^{\mathrm{\prime }}$, absolutely continuous on $[a,b]$, with $w(x)-w(y)={\int }_y^x{D}^{2}u$ for all $x,y\in [a,b]$. The a.e. derivative of $u^{\ast }$ equals the class $u^{\mathrm{\prime }}$, so it equals $w$ almost everywhere. The fundamental theorem for absolutely continuous functions therefore gives $u^{\ast }(x)=u^{\ast }(a)+{\int }_a^{x}w(t)dt$ for every $x\in [a,b]$. Since $w$ is continuous, this identity makes $u^{\ast }$ continuously differentiable and gives $(u^{\ast }{)}^{\mathrm{\prime }}(x)=w(x)$ at every $x\in [a,b]$. Thus $(u^{\ast }{)}^{\mathrm{\prime }}$ is absolutely continuous and has the displayed difference formula, proving (c).