I know this is a special case of Sobolev embedding theorem but I heard there is a simple way to prove this special case. Seems to start with the dense subset $C^\infty [a,b]$. Construct a Cauchy sequence for any function in $H^1$. I'm lost as to what to do next?
Someone suggest that $\forall v\in H^1\cap C^\infty$, $|v(x)-v(y)|=|\int_x^y{v'(t)}dt| \leq |\int_x^y{1^2}dt||\int_x^y{v'(t)^2}dt| \leq \sqrt{y-x}\| v\|_1$. This imply $v$ is Holder continuous, but is that true for general $H^1$ function?
And that is not equicontinuity, $\forall \epsilon ,\exists \delta,\forall x,y \text{ s.t. } |x-y|<\delta,|v(x)-v(y)|<\epsilon ,\forall v$ ,$\delta$ should solely depend on $\epsilon$. However, in our case, $\delta$ has to rely on choice of function.
Given: a sequence of smooth functions $v_j$ that is convergent in $H^1$.
Goal: show that $v_j$ converge in $C^0$.
Method: prove that $\|v_j-v_k\|_{C^0}$ is small when $\|v_j-v_k\|_{H^1}$ is small.
Details: let $u=v_j-v_k$. Note that $|\int_a^b u|\le \sqrt{b-a}\sqrt{\int_a^b u_j^2}\le \sqrt{b-a}\|u\|_{H^1}$. By the mean value theorem for integrals, $u$ attains the value $m=\frac{1}{b-a}\int_a^b u$ at some point $x_0$. Then your continuity estimate gives $$|u(x)-m|\le \sqrt{|x-x_0|}\|u\|_{H^1}\le \sqrt{b-a}\|u\|_{H^1}$$ for all $x\in [a,b]$. Thus, $$ \|u\|_{C^0}\le m+ \sqrt{b-a}\|u\|_{H^1} \le \frac{1}{\sqrt{b-a}}\|u\|_{H^1}+ \sqrt{b-a}\|u\|_{H^1} $$
Mission accomplished: the sequence $v_j$ is Cauchy in $C^0[a,b]$, and therefore converges there.