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.
**Added:** In light of everything said, the easiest approach seems to be to just take $f\_\epsilon(x) = \eta(x/\epsilon) e^{x-1}$, where $\eta$ is a smooth function with $\eta(0) = 0$ and $\eta(x) = 1$ for $x\geq 1$ (so that $\int\_0^1 \eta'= 1$). The integral is then $e^{-1}\int\_0^\epsilon \epsilon^{-1}\eta'(x/\epsilon)e^x\,dx = e^{-1}\int\_0^1 \eta'(x) e^{\epsilon x}\,dx$, which tends to $e^{-1}$ as $\epsilon\to0$. I blame the desire to write things down in terms of elementary functions for the over-complication below.

---

Here's an idea for a sequence of minimizers. (As the comment points out your proof shows there's no actual minimizer.) Take $f\_\alpha(x) = x^\alpha e^{x-1}$ and let $\alpha \to 0$; the point is to make the function look more and more like an exponential as the parameter is varied. Note that $f\_\alpha'(x) - f\_\alpha(x) = \alpha x^{\alpha - 1} e^{x-1}$, and making the substitution $y = x^\alpha$ in the integral should show that the integrals tend to $e^{-1}$ as $\alpha \to 0$.

**Edit:** John pointed out that $f\_\alpha$ isn't differentiable at $0$. To fix this, take some smooth approximation to $f\_\alpha$ instead, where the approximation becomes better (sufficiently fast) as the parameter tends to $0$. To be explicit, you could use $g\_\alpha = \eta\_\alpha f\_\alpha$, where $\eta\_\alpha$ is smooth and increasing with $\eta\_\alpha(0) = 0$ and $\eta\_\alpha(\epsilon) = 1$ for some small $\epsilon$ (depending on $\alpha$). For the approximation to work you'll just need to make sure that $\int\_0^\epsilon \eta\_\alpha'(x) x^\alpha \to 0$ with $\alpha$. You can arrange for $\eta\_\alpha'\lesssim 1/\epsilon$, and then the integral will be $\lesssim\epsilon^\alpha$, so, for instance, $\epsilon = \epsilon(\alpha) = \alpha^{1/\alpha}$ should do.