Let $f,g \in \mathcal C^0([0,1],\mathbb R)$ and $L \ge 1$. We endow $\mathcal C^0([0,1],\mathbb R)$ with the sup norm $\|\cdot\|_\infty$. Consider $$\Phi: [0,1] \to \mathbb R, \quad x \mapsto L e^{-2Lx} \int_0^x | f(t)- g(t) | \, \mathrm{d} t$$ Find the maximum of $\Phi$.
In order to find the maximum, I take the derivative of $\Phi$ and compute the values of $\Phi$ at the critical points and at $0$ and $1$. After that. I'm unable to compare those values. Could you please shed me some light how to finish the task? Thank you so much!
Because $\Phi \in \mathcal C^1([0,1],\mathbb R)$, the maximum is attained at critical points or at $0$ or at $1$. We have $$\begin{aligned} \Phi'(x) &= -2L^2 e^{-2Lx} \int_0^x | f(t)- g(t) | \, \mathrm{d} t + L e^{-2Lx} | f(x)- g(x) | \\ &= L e^{-2Lx} \left (| f(x)- g(x) | - 2L \int_0^x | f(t)- g(t) | \, \mathrm{d} t \right )\end{aligned}$$
If $\bar x$ is a critical point, then $\Phi'(\bar x) = 0$ and consequently $$\frac{1}{2} | f(\bar x)- g(\bar x) | = L \int_0^{\bar x} | f(t)- g(t) | \, \mathrm{d} t$$
Then $$\Phi(\bar x) = \frac{1}{2} e^{-2L \bar x} | f(\bar x)- g(\bar x) |$$
It follows that $\Phi (0)=0$ and $$\Phi(\bar x)= \frac{1}{2} e^{-2L \bar x} | f(\bar x)- g(\bar x) |$$ and $$\Phi(1) = L e^{-2L} \int_0^1 | f(t)- g(t) | \, \mathrm{d} t$$