🏗️ $\Theta \rho ϵ\eta \Pi \alpha \tau \pi$🚧 (under construction)

Note that if $0<a<b$ then $${\displaystyle {\int }_a^{b}sign\left(x\right)dx={\int }_a^{b}1dx=b-a=\left|b\right|-\left|\right|}$$ when $0>b>a$ then we have $${\displaystyle {\int }_a^{b}sign\left(x\right)dx={\int }_a^b-1dx=a-b==-b-(-a)=\left|b\right|-\left|a\right|}$$ and finally when $a<0<b$ then we have $${\displaystyle {\int }_a^{b}sign\left(x\right)dx={\int }_a^0-1dx+{\int }_0^{b}1dx=a+b=b-(-a)=\left|b\right|-\left|a\right|}$$ which shows that ${\int }_a^{b}f\left(x\right)dx=F\left(b\right)-F\left(a\right)$ in all cases, note that $F$ is not an antiderivative of $f$ because it is undefined at some points in the domain of $f$.

Let's try to compute $G^{\prime }$, if it turns out that by observation that equation we get for $G^{\prime }$ is continous then $G$ is $C^1$, we approach by definition: $${\displaystyle G^{\prime }\left(x\right)={lim}_{h\to 0}\frac{{\int }_{x+h}^{x+h+ϵ}f\left(t\right)dt-{\int }_x^{x+ϵ}f\left(t\right)dt}{h}}$$ Since $h\to 0$ we can assume that eventually $h<ϵ$ so that $${\displaystyle x<x+h<x+ϵ<x+h+ϵ}$$ therefore the integrals have an overlapping cancellation from $x+h$ to $x+ϵ$ so that $${\displaystyle ϵ·G^{\prime }\left(x\right)={lim}_{h\to \infty }\frac{{\int }_{x+ϵ}^{x+h+ϵ}f\left(t\right)dt-{\int }_x^{x+h}f\left(t\right)dt}{h}}$$ Let $a\in \mathbb{R}$ and define $K\left(b\right)={\int }_a^{b}f\left(t\right)dt$ so that we have $${\displaystyle {\int }_{x+ϵ}^{x+h+ϵ}f\left(t\right)dt=-{\int }_a^{x+ϵ}f\left(t\right)dt+{\int }_a^{x+h+ϵ}f\left(t\right)dt=K(x+h+ϵ)-K(x+ϵ)}$$ and also $${\displaystyle {\int }_x^{x+h}=-{\int }_a^{x}f\left(t\right)dt+{\int }_a^{x+h}f\left(t\right)dt=K(x+h)-K\left(x\right)}$$ Therefore by finally constructing $J\left(p\right)=K(p+ϵ)-K\left(p\right)$ we have $$\begin{gathered} {\displaystyle ϵ·G^{\prime }\left(x\right)\&={lim}_{h\to \infty }\frac{{\int }_{x+ϵ}^{x+h+ϵ}f\left(t\right)dt-{\int }_x^{x+h}f\left(t\right)dt}{h} \\ \&={lim}_{h\to \infty }\frac{(K(x+h+ϵ)-K(x+ϵ))-(K(x+h)-K\left(x\right))}{h} \\ \&={lim}_{h\to \infty }\frac{(K(x+h+ϵ)-K(x+h))-(K(x+ϵ)-K\left(x\right))}{h} \\ \&={lim}_{h\to \infty }\frac{J(x+h)-J\left(x\right)}{h} \\ \&=J^{\prime }\left(x\right) \\ \&=K^{\prime }(x+ϵ)-K^{\prime }\left(x\right) \\ \&=f(x+ϵ)-f\left(x\right) \\ } \end{gathered}$$ therefore $${\displaystyle G^{\prime }\left(x\right)=\frac{f(x+ϵ)-f\left(x\right)}{ϵ}}$$ as the right hand side is composition of fundamental operations on $\mathbb{R}$ which maintain continuity then $G^{\prime }\left(x\right)$ is continuous, so that $G$ is $C^1$ as needed.

Let $x_0\in \mathbb{R}$, and let $f^{\prime }\left(x_0\right)=b$, if it's the case that $b>0$, then let $t\in \mathbb{R}$ if $t>0$ then we apply MVT on the interval $[x_0,x_0+t]$ to get some ${\theta }_1\in (x_0,x_0+t)$ such that $${\displaystyle \frac{f^{\prime }(x_0+t)-f^{\prime }\left(x_0\right)}{t}=f^{\prime \prime }\left({\theta }_1\right)}$$ so that $${\displaystyle f^{\prime }(x_0+t)=f^{\prime \prime }\left({\theta }_1\right)t+b}$$ and that $${\displaystyle f^{\prime }(x_0+t)\ge b-Ct=b-C\left|t\right|}$$ Now if $t<0$ one can do the same analysis but will obtain some ${\theta }_2\in [x_0+t,x_0]$ such that $${\displaystyle \frac{f^{\prime }\left(x_0\right)-f^{\prime }(x_0+t)}{-t}=f^{\prime \prime }\left({\theta }_2\right)}$$ so that $${\displaystyle f^{\prime }(x_0+t)=f^{\prime }\left(x_0\right)-f^{\prime \prime }\left({\theta }_2\right)(-t)=b-f^{\prime \prime }\left({\theta }_2\right)\left|t\right|\ge b-C\left|t\right|}$$ Now when $x_0=0$, we have the following integral inequalities, note the the first integral exists because $f^{\prime }$ is continuous: $${\displaystyle {\int }_{-\frac{b}{C}}^{\frac{b}{C}}{f}^{\prime }\left(t\right)dt\ge {\int }_{-\frac{b}{C}}^{\frac{b}{C}}b-C\left|t\right|dt}$$ But note that the integral on the right measure the area of the function $b-C\left|t\right|$ which hits the y axis at $b$ and has legs coming down which intersect the $x$ axis at $-\frac{b}{C},\frac{b}{C}$ respectively, so that the total area is given by $\left(\frac{b}{C}\right)b=\frac{b^2}{C}$. Now focusing on the left hand side we recall that since $f$ is a continuous function such that $f^{\prime }\left(x\right)=f^{\prime }\left(x\right)$ then we have: $${\displaystyle {\int }_{-\frac{b}{C}}^{\frac{b}{C}}{f}^{\prime }\left(t\right)dt=f\left(\frac{b}{C}\right)-f(-\frac{b}{C})\le 2A}$$ thus chaining previous inqualities with the above equality we have: $${\displaystyle b\le \sqrt{2AC}}$$ If it turns out that $b$ is negative we can get the same bound, which shows for an arbitrary choice of $x_0$ we were able to deduce that $\left|f\left(x_0\right)\right|\le \sqrt{2AC}$, thus $${\displaystyle \Vert f^{\prime }{\Vert }_{\infty }\le \sqrt{2AC}}$$