Length-energy inequality and constant-speed equality case

Put $\mathrm{\ell }=b-a>0$ and $c=L\mathrm{/}\mathrm{\ell }$. Finite additivity and ordinary integral algebra on the smooth pieces give $0\le {\int }_a^b(h-c{)}^2={\int }_a^b{h}^2-2c{\int }_a^{b}h+c^2\mathrm{\ell }=2E-L^2\mathrm{/}\mathrm{\ell }\mathrm{.}$ Multiplying by $\mathrm{\ell }$ proves $L^2\le 2\mathrm{\ell }E$.

Conversely, if $h=c_0$ almost everywhere, changing finitely many breakpoint values does not affect either integral, so $L=c_0\mathrm{\ell }$ and $2E=c_0^2\mathrm{\ell }$. Hence $L^2=c_0^2{\mathrm{\ell }}^2=2\mathrm{\ell }E$. The same computation applies when $c_0=0$.

If equality holds, step 1.1 gives a zero total integral of $(h-c{)}^2$. Every piece integral is nonnegative, so each is zero. On the interior of each piece, $(h-c{)}^2$ is continuous; applying [F2] on compact subintervals contained in that interior gives $h=c$ there. Thus $h=c$ away from the finitely many breakpoints, hence almost everywhere, and it has the same constant value on every smooth piece.

Steps 2.1 and 1.2 prove both directions of the equality characterization. A constant curve has $L=E=0$ and realizes equality. In dimension zero every curve is locally constant, so the same case applies; dimension one is unchanged. A curve on a nonempty interval cannot have empty target. The hypothesis $a<b$ excludes division by a zero interval length; endpoint and corner values occur on a finite set and do not alter the integrals. No point or representative is chosen from an indexed family, so the proof is choice-free.