# 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.

• You want to prove equicontinuity of functions with bounded $H^1$ norm. And Hölder continuity in the form you have implies that.
– user53153
Feb 6, 2013 at 3:57
• Read my comment again: for any bounded family of functions in $H^1$ ($\|v\|_{H^1}\le M$) you have equicontinuity by means of $|v(x)-v(y)|\le M\sqrt{x-y}$. This is all you need, because convergent sequences in $H^1$ are bounded.
– user53153
Feb 6, 2013 at 4:23
• I think "uniformly bounded" in Arzela-Ascoli theorem is with regard to $C^0$ norm, not $H^1$ norm. Feb 6, 2013 at 4:41

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.

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 $$H^1[a,b] \subset C[a,b]$$