Question: 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 $\int\_0^1|f'(x)-f(x)|dx=1/e$?

Answer: 
Since
$$
\int\_0^1(f'(x)-f(x))\,e^{-x}\,\mathrm{d}x=\frac{f(1)}e-f(0)\tag{1}
$$
we have
$$
\int\_0^1(f'(x)-f(x))\,\mathrm{d}x-\left(\frac{f(1)}e-f(0)\right)=\int\_0^1(f'(x)-f(x))\left(1-e^{-x}\right)\mathrm{d}x\tag{2}
$$
To make $(2)$ as small as possible, we would like to concentrate $f'(x)-f(x)$ where $1-e^{-x}$ is small; that is, near $x=0$.

Consider $f\_a(x)=x^ae^{x-1}$ for $a\gt0$. We get $f\_a'(x)-f\_a(x)=ax^{a-1}e^{x-1}$.
$$
\begin{align}
\lim\_{a\to0^+}\int\_0^1|f\_a'(x)-f\_a(x)|\,\mathrm{d}x
&=\lim\_{a\to0^+}\int\_0^1ax^{a-1}e^{x-1}\,\mathrm{d}x\\
&=\lim\_{a\to0^+}\left(\left[x^ae^{x-1}\vphantom{\int}\right]\_0^1-\int\_0^1x^ae^{x-1}\,\mathrm{d}x\right)\\
&=\lim\_{a\to0^+}\left(1-\int\_0^1x^ae^{x-1}\,\mathrm{d}x\right)\\
&=1-\int\_0^1e^{x-1}\,\mathrm{d}x+\lim\_{a\to0^+}\left(\int\_0^1\left(1-x^a\right)e^{x-1}\,\mathrm{d}x\right)\\[6pt]
&=e^{-1}\tag{3}
\end{align}
$$
Since
$$
0\le\int\_0^1\left(1-x^a\right)e^{x-1}\,\mathrm{d}x\le\int\_0^1\left(1-x^a\right)\,\mathrm{d}x=\frac{a}{1+a}\tag{4}
$$
the Squeeze Theorem says that
$$
\lim\_{a\to0^+}\left(\int\_0^1\left(1-x^a\right)e^{x-1}\,\mathrm{d}x\right)=0\tag{5}
$$
Thus, $(3)$ implies the bound is sharp.
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)).