$f'$ is in $L^2[0,1]$ Let $f$ is absolutely continuous function on $[0,1]$, $f(0)=0$ and $f' \in L^2[0,1]$. Would you help me to prove that there is constant $c$ such that
$$|f(t)| \leq c \left( \int_0^1 |f'(t)|^2 dt \right)^{1/2}$$
 A: Since 
$$
f(t) = f(0) + \int_0^t f'(x) dx  = \int_0^t f'(x) dx 
$$
we have
$$
\lvert f(t) \rvert \leq \int_0^t \lvert f'(x) \rvert dx \leq \int_0^1 1 \cdot\lvert f'(x) \rvert dx
$$
Now, using Cauchy-Bunyakovsky-Schwarz inequality in $L^2[0, 1]$, we conclude
$$
\int_0^1 1 \cdot\lvert f'(x) \rvert dx\leq \left(\int_0^1 1^2\cdot dx \right)^{1/2} \left(\int_0^1 \lvert f'(x) \rvert^2 dx \right)^{1/2} = \left(\int_0^1 \lvert f'(x) \rvert^2 dx \right)^{1/2}
$$
A: The total variation of $f\in W^{1,1}[0,1]$ is given by
$$
\operatorname*{Var}_{[0,1]}\,f=\int_0^1\left|f^\prime(t)\right|\,\mathrm{d}t\tag{1}
$$
$(1)$ implies that for any $x,y\in[0,1]$,
$$
\left|f(x)-f(y)\right|\le\int_0^1\left|f^\prime(t)\right|\,\mathrm{d}t\tag{2}
$$
Furthermore, for any convex $\phi$, Jensen's Inequality says
$$
\phi\left(\int_0^1g(t)\,\mathrm{d}t\right)\le\int_0^1\phi(g(t))\,\mathrm{d}t\tag{3}
$$
Using the special cases of $y=0$, $g=\left|f'\right|$, and $\phi(x)=x^2$ yields
$$
\left|f(x)\right|^2\le\int_0^1\left|f^\prime(t)\right|^2\,\mathrm{d}t\tag{4}
$$
