Show $\inf_f\int_0^1|f'(x)-f(x)|dx=1/e$ for continuously differentiable functions with $f(0)=0$, $f(1)=1$. Let $C$ be the class of all real-valued continuously differentiable functions $f$ on the interval $[0,1]$ with $f(0)=0$ and $f(1)=1$. How to show that
$$\inf_{f\in C}\int_0^1|f'(x)-f(x)|dx=\frac{1}{e}?$$
I have been able to show that $1/e$ is a lower bound. Indeed,
$$\begin{align*}
\int_0^1|f'(x)-f(x)|dx &= \int_0^1|f'(x)e^{-x}-f(x)e^{-x}|e^xdx \\
&\geq \int_0^1\left(f'(x)e^{-x}-f(x)e^{-x}\right) dx \\
&= \int_0^1 \frac{d\left(f(x)e^{-x}\right)}{dx}dx  \\
&= f(1)e^{-1}-f(0)e^{0}\\
&=\frac{1}{e}.
\end{align*}$$
But how to show this is the infimum? Is there a function $f\in C$ such that we get $\int_0^1|f'(x)-f(x)|dx=1/e$?

Here is another method.
Let $f \in C^1([0,1])$ with $f(0)=0$ and $f(1)=1$, and $g=f'-f$.
then solving the ODE : $f' =f +g$ with $f(0)=0$ gives
$$f(x) = e^x \int_0^x  e^{-t} g(t) dt.$$
Since $f(1)=1$, we get
$$ \int_0^1 e^{-t} g(t) dt = \frac{1}{e} \quad \quad (1). $$
By Holder we have : $| \int_0^1 e^{-t} g(t) dt| \leq \|g\|_1$.
Hence
$$\|g\|_1 = \int_0^1 |f'-f| \geq 1/e,$$
and equality holds iff $g$ goes to the dirac $1/e.\delta_0$ (with the condition (1)).
