Why is $H^1[a,b]\subset C^0[a,b]$? 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.
 A: 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.
A: Cauchy sequence $\{v_n\}$ in $H^1[a,b] \cap C^{\infty}$[a,b] with norm  $||\cdot||_{H_1}$, we have for any small $\epsilon > 0$, there exists p $\in N^*$, such that $||v_n - v_m||_{H_1} < \epsilon$, for any $m, n > p$. Let $\epsilon = \epsilon_0$,  concide with $ p = p_0(\epsilon_0)$, and fix $m = m_0 > p_0$, we have $||v_n||_{H_1} \leq ||v_n - v_{m_0}||_{H_1}  + ||v_{m_0}||_{H_1} < \epsilon_0 + M_0 = M$, where $M_0 = \sum_{i = 0} ^ {i = m_0} ||v_{i}||_{H_1}   $ for any $n > p$.  Also $||v_n||_{H_1}  < M$ for any $n \in N$.
So  $\{v_n\}$ is bounded in norm $H_1$.
Next prove equicontinuous.
For any $n \in N^*$, and any small $\epsilon > 0$, there exists $\delta = (\epsilon /M)^{1/2}$, such that $|v_n(x) - v_n(y)|  = |\int_x ^y Dv_n(t) dt | \leq |\int_x ^y 1 ^2 dt |^{1/2} \cdot |\int_x ^y Dv_n(t) ^ 2 dt |^{1/2}\leq \sqrt{\delta} ||v_n||_{H_1} < \sqrt{\delta}  M = \epsilon$ , where $|x-y| <\delta = (\epsilon /M)^{1/2}$.
$\{v_n\}$ is equicontinuous and bounded. As by theorem of Arzela-Ascoli $\{v_n\}$ is uniformly convergent, we get that the limiting function $v$ is continuous. 
Any function $v \in H^1[a,b]$, we can find Cauchy sequence in $H^1[a,b]\cap C^{\infty}[a,b]$, $\{v_n\}$ converges to v, $(C^{\infty}[a,b]$ is dense in $H^1[a,b]$). And $v$ is continous. 
So $H^1[a,b] \subset C[a,b]$
