Do we necessarily have that $W^{2, p}(I) \subset C^1(\overline{I})$ with compact injection?

Question

Let $I = (0, 1)$ and $p > 1$. Do we necessarily have that$$W^{2, p}(I) \subset C^1(\overline{I})$$with compact injection?

Answer 1

Yes. In fact, we have the following more general result.

Theorem (Adams, page 168). Let $\Omega$ be a domain in $\mathbb R^n$ and $\Omega_0$ be a bounded subdomain of $\Omega$. Let $j \geq 0$ and $m \geq 1$ be integers, and let $ 1 \leq p < \infty$. If $\Omega$ satisfies the strong local Lipschitz condition, then the following embedding is compact: $$W^{j+m,p}(\Omega)\subset C^j(\overline{\Omega_0}),\qquad \text{if}\quad mp>n.$$

Notes:

• If $\Omega$ is bounded, we may have $\Omega_0=\Omega$ in the statement of the theorem.
• The bounded subdomain $\Omega_0$ of $\Omega$ may be assumed to satisfy the cone condition if $\Omega$ does.

---

The proof of your particular case, can be done as follows.

Fact 1. $W^{2,p}(I)\subset C^1(\overline{I})$

Proof: Take $u\in W^{2,p}(0,1)$. Define $$h(x)=\int_0^xu''(t)\;dt,\qquad g(x)=\int_0^xu'(t)\;dt.$$ Then, $$\int_0^1(u'-h)\varphi'=\int_0^1u'\varphi'-\int_0^1h\varphi'=-\int_0^1u''\varphi+\int_0^1u''\varphi=0,\quad\forall\ \varphi\in C^1_c(0,1)$$ and thus $u'-h=c$ for some constant $c$. This imply $u'=h+c\in C(\overline{I})$. Analogously, there is a constant $c_2$ such that $u=g+c_2\in C(\overline{I})$. So, $W^{2,p}(I)\subset C^1(\overline{I})$.

Fact 2. The inclusion $W^{2,p}(I)\subset C^1(\overline{I})$ is continuous.

Proof: Take $u\in W^{2,p}(I)$ and $x\in I$. Then, $$u'(x)=u'(0)+\int_0^xu''(t)\;dt$$ and thus $$|u'(x)|\leq |u'(0)|+\|u''\|_{L^1},\quad |u'(0)|\leq |u'(x)|+\|u''\|_{L^1}.\tag{1}$$ From the second inequality in $(1)$, $$|u'(0)|=\int_0^1|u'(0)|\;dt\leq \|u'\|_{L^1}+\|u''\|_{L^1}$$ So, from the first inequality in $(1)$, $$|u'(x)|\leq \|u'\|_{L^1}+2\|u''\|_{L^1}\leq c_0\|u\|_{W^{2,p}}$$ for some constant $c_0$. A similar argument shows that $$|u(x)|\leq \|u\|_{L^1}+2\|u'\|_{L^1}\leq c_0\|u\|_{W^{2,p}}.$$ As $x\in I$ is arbitrary, we get $$\|u\|_{C^1}=\sup_{x\in\overline{I}} |u(x)|+\sup_{x\in\overline{I}} |u'(x)|\leq 2 c_0\|u\|_{W^{2,p}}.$$

Fact 3. The inclusion $W^{2,p}(I)\subset C^1(\overline{I})$ is compact.

Proof: Let $(u_n)$ be a bounded sequence in $W^{2,p}(0,1)$, say by a constant $M$.

The sequence $(u'_n)$ is uniformly bounded because, from the proof of the Fact 2, $|u'_n(x)|\leq c_0 M$ for all $n\in\mathbb{N}$ and all $x\in \overline{I}$. Also, $(u'_n)$ is equicontinuous because, from the Hölder's inequality, $$|u'_n(x)-u'_n(y)|=\left|\int_y^xu_n''(t)\;dt\right|\leq \|u''\|_{L^p}\left(\int_y^x1\;dt\right)^{\frac{1}{p}}\leq M|x-y|^{1/p}$$ for all $n\in\mathbb{N}$ and all $x,y\in \overline{I}$. So, the Arzelà–Ascoli Theorem implies that $(u'_n)$ has a subsequence $(u'_{n_k})$ that converges uniformly to some $g\in C(\overline{I})$. A similar argument shows that $(u_{n_k})$ has a subsequence $(u_{n_{k_m}})$ that converges uniformly to some $u\in C(\overline{I})$.

It follows from Theorem 7.17 in Rudin's book that $u'=g$ and thus $(u_n)$ has a subsequence $(u_{n_{k_m}})$ that converges in $C^1(\overline{I})$.