Descent inequality for a Lipschitz gradient

Put $d=y-x$. If $d=0$ the inequality is equality. Otherwise define $\varphi (s)=f(x+sd)$ on $[0,1]$. By [F2] and continuity of $\mathrm{\nabla }f$, $\varphi$ is continuous on $[0,1]$, differentiable on $(0,1)$, and ${\varphi }^{\mathrm{\prime }}(s)=⟨\mathrm{\nabla }f(x+sd),d⟩$.

For an arbitrary natural $N\ge 1$, partition $[0,1]$ at $j\mathrm{/}N$. Applying [F1] on each subinterval gives ${\xi }_j\in (j\mathrm{/}N,(j+1)\mathrm{/}N)$ with $\varphi ((j+1)\mathrm{/}N)-\varphi (j\mathrm{/}N)={\varphi }^{\mathrm{\prime }}({\xi }_j)\mathrm{/}N$. Telescoping and subtracting ${\varphi }^{\mathrm{\prime }}(0)$ yields $f(y)-f(x)-⟨\mathrm{\nabla }f(x),d⟩=\frac{1}{N}{\sum }_{j=0}^{N-1}({\varphi }^{\mathrm{\prime }}({\xi }_j)-{\varphi }^{\mathrm{\prime }}(0))\le \frac{L\mathrm{\parallel }d{\mathrm{\parallel }}_2^2}{N^2}{\sum }_{j=0}^{N-1}(j+1)=\frac{L}{2}(1+1\mathrm{/}N)\mathrm{\parallel }d{\mathrm{\parallel }}_2^2,$ because [F3] and [F4] give $\mathrm{\mid }{\varphi }^{\mathrm{\prime }}(s)-{\varphi }^{\mathrm{\prime }}(0)\mathrm{\mid }\le Ls\mathrm{\parallel }d{\mathrm{\parallel }}_2^2$ and ${\xi }_j<(j+1)\mathrm{/}N$. To evaluate the finite sum for every natural $N$, its value is $N(N+1)\mathrm{/}2$: at $N=0$ both sides are zero, and the recursion [F6] adds $N+1$ when $N$ is increased by one, giving $(N+1)(N+2)\mathrm{/}2$. [F7] proves this identity by induction. Only finitely many mean-value points are used for each fixed $N$.

The left side of the bound in step 1.2 is independent of $N$. Suppose it exceeded $L\mathrm{\parallel }d{\mathrm{\parallel }}_2^2\mathrm{/}2$ by $\delta >0$. For $L\mathrm{\parallel }d{\mathrm{\parallel }}_2^2>0$, [F5] lets us choose $N$ with $1\mathrm{/}N<2\delta \mathrm{/}(L\mathrm{\parallel }d{\mathrm{\parallel }}_2^2)$, making the right side in step 1.2 strictly smaller than the left side, a contradiction. If $L\mathrm{\parallel }d{\mathrm{\parallel }}_2^2=0$, step 1.2 already gives the desired bound.