There is a paper in which the authors state that, if we consider the Sobolev space $$ \tilde{H}^1 = \{ u \in H^1(0,l)\subseteq\mathbb{R} : u(0) = 0 \} $$ the standard norm $$ \|u\|_{H^1} = \sqrt{\int_0^l\big\{[u(x)]^2+[u'(x)]^2\big\}dx} $$ is equivalent to the following norm $$ \|u\| = \sqrt{\int_0^l[u'(x)]^2dx} $$ This means that two positive constants $C_1$, $C_2$ exist such that $$ C_1\sqrt{\int_0^l[u'(x)]^2dx} \leq \sqrt{\int_0^l\big\{[u(x)]^2+[u'(x)]^2\big\}dx} \leq C_2\sqrt{\int_0^l[u'(x)]^2dx} $$ While the first inequality is obviously true and the second one would be obviously false without the condition $u(0)=0$, I cannot found a proof for the second one, if true, or a counterexample, if false.
I tried with $u_k(x)=x^k$, looking for
$$
\sup_{k\in\mathbb{N}}\left\{\frac{\|u_k\|_{H^1}}{\|u_k\|}\right\}
$$
and with
$$
u_h(x)=\begin{cases}
mx & 0 \leq x \leq h \\
mh & h < x \leq 1
\end{cases}
$$
for $0<h<1$, but they are not counterexamples, given that the $\sup$ exists.
I found the Wirtinger's inequality for functions, but does not seem to help.
Edit
Can the Cantor function serve as a counterexample?
So first, the Cantor function is not a counterexample as its derivative (in the sense of distributions) is a measure that is not absolutely continuous with respect to the Lebesgue measure, and so in particular is not in $L^2$.
The property your are looking at is indeed close to the Poincaré$-$Wirtinger inequality and can be proved similarly, for example, using the usual reasoning by contradiction. Let me give you a direct and quantitative proof.
Since $u(0) = 0$, it follows that for any $y\in(0,l)$, $$ u(y) = \int_0^y u'(x)\,\mathrm d x $$ hence taking the square, integrating in $y$ and using the Cauchy-Schwarz inequality, $$ \int_0^l |u(y)|^2\,\mathrm d y = \int_0^l\left|\int_0^y u'(x)\,\mathrm d x\right|^2\,\mathrm d y \\ \leq \int_0^l y\int_0^y |u'(x)|^2\,\mathrm d x\,\mathrm d y \\ \leq \int_0^l y \,\mathrm d y \int_0^l |u'(x)|^2\,\mathrm d x = \frac{l^2}{2} \int_0^l |u'(x)|^2\,\mathrm d x $$ and so $$ \int_0^l |u|^2 + |u'|^2 \leq \left(\frac{l^2}{2}+1\right) \int_0^l |u'|^2. $$