Equivalence of Sobolev norm with infinity norm Let $f\in C^1[a,b]$ be a function. The Sobolev norm is defined as follows:
$$\|f\|_{1,2}=\left[\int_a^b \left(f(x)^2 +f'(x)^2\right)dx\right]^{1/2}$$
I want to find a constant $C$ such that $\|f\|_\infty\le C\|f\|_{1,2}$ for all $C^1[a,b]$.
How do I do so? I've tried some approaches with inequalities, but didn't succeed. Would appreciate a hint.
Also, does the existence of such a constant imply that the norms are equivalent? (Of course, we also need to find $c$ so that $\|f\|_\infty\ge c\|f\|_{1,2}$, but just curious. Since $(C^1[a,b], \|\cdot\|_\infty$ is complete and $(C^1[a,b], \|\cdot\|_{1,2})$ is incomplete, can the norms be possibly equivalent?

Let $f\in E={\mathcal C}^1([a,b])$. Then there exists $m\in I=[a,b]$ such that $|f(t)|\geq |f(m)|$ for all $t\in I$. We have also $$f(x)=\int_m ^x f^{\prime}(t)dt+f(m)$$
hence $$|f(x)|\leq |f(m)|+\int_{a}^{b}|f^{\prime}(t)|dt\leq \frac{1}{b-a}\int_a^b|f(t)|dt+\int_{a}^{b}|f^{\prime}(t)|dt$$
By Cauchy-Schwarz, we have $(\int_a^b|f(t)|dt)^2\leq (b-a)\int_a^b|f(t)|^2dt$ and $(\int_{a}^{b}|f^{\prime}(t)|dt)^2\leq (b-a)\int_a^b|f^{\prime}(t)|^2dt$, and so there exists a constant $c_1>0$ such that $$|f(x)|\leq c_1((\int_a^b|f(t)|^2dt)^{1/2}+(\int_a^b|f^{\prime}(t)|^2dt)^{1/2}) $$
for all $x\in I$. To finish, use that if $u,v\geq 0$, we have $u+v\leq \sqrt{2}\sqrt{u^2+v^2}$. 
