Exists $C = C(\epsilon, p)$ where $\|u\|_{L^\infty(0, 1)} \le \epsilon\|u'\|_{L^p(0, 1)} + C\|u\|_{L^1(0, 1)}$ for all $u \in W^{1, p}(0, 1)$? Let $p > 1$. For all $\epsilon > 0$, does there exist $C = C(\epsilon, p)$ such that$$\|u\|_{L^\infty(0, 1)} \le \epsilon\|u'\|_{L^p(0, 1)} + C\|u\|_{L^1(0, 1)}$$for all $u \in W^{1, p}(0, 1)$?
 A: Yes and to prove it it's enough to prove that for all $\delta>0$, there exists $K=K(\delta,p)$ such that
$$\|u\|_{L^\infty(0, 1)} \le \delta\|u\|_{W^{1,p}(0, 1)} + K\|u\|_{L^1(0, 1)}, \quad\forall\ u \in W^{1, p}(0, 1).\tag{1}$$
In fact, assume that $(1)$ is true. Let $c>0$ be a constant such that $\|\cdot\|_{L^p}\leq c\|\cdot\|_{L^\infty}$. Given $\varepsilon>0$, take $\delta=\frac{\varepsilon}{1+c\varepsilon}$. Then
\begin{align}\|u\|_{L^\infty(0, 1)} &\le \delta\|u'\|_{L^p(0, 1)} + \delta\|u\|_{L^p(0, 1)}+K\|u\|_{L^1(0, 1)}\\
&\le \delta\|u'\|_{L^p(0, 1)} + \delta c \|u\|_{L^\infty(0, 1)}+K\|u\|_{L^1(0, 1)}
\end{align}
which implies
\begin{align}(1-c\delta)\|u\|_{L^\infty(0, 1)} &\le \delta\|u'\|_{L^p(0, 1)} +K\|u\|_{L^1(0, 1)}
\end{align}
Thus the desired result holds with $C=K(1+c\varepsilon)$.

Proof of $(1)$: Assume that $(1)$ is not true. Then, there exists a constant $\eta_0>0$ and a sequence $(u_n)$ in $W^{1,p}(0,1)$ such that
$$\|u_n\|_{L^\infty}>\eta_0\|u_n\|_{W^{1,p}}+n\|u_n\|_{L^1},\quad\forall\ n\in \mathbb{N}.$$
Note that $u_n\neq 0$ for all $n\in\mathbb{N}$ (otherwise we would have $0>0$) and thus we can define
$$w_n=\frac{u_n}{\|u_n\|_{W^{1,p}}}$$
which satisfies
$$\|w_n\|_{L^\infty}>\eta_0+n\|w_n\|_{L^1},\quad\forall\ n\in \mathbb{N}.\tag{1}$$
On the other hand, as the inclusion $W^{1,p}\subset L^\infty$ is continuous, there exists a constant $L>0$ such that
$$\|w_n\|_{L^\infty}\leq L\|w_n\|_{W^{1,p}}=L,\quad \forall\ n\in\mathbb{N}.\tag{2}$$
It follows from $(1)$ and $(2)$ that
$$\|w_n\|_{L^1}<\frac{L}{n},\quad\forall\ n\in\mathbb{N}.\tag{3}$$
As the inclusion $W^{1,p}\subset L^\infty$ is compact (because $p>1$), there exist $w_0\in L^\infty$ and a subsequence of $(w_n)$, which will not be relabeled, such that
$$\|w_n- w_0\|_{L^\infty}\to 0.\tag{4}$$
As the inclusion $L^\infty\subset L^1$ is continuous, it follows that
$$\|w_n- w_0\|_{L^1}\to 0.\tag{5}$$
From $(3)$ and $(5)$, we get $w_0=0$. So, from $(1)$ and $(4)$,
$$\|w_n\|_{L^\infty}>\eta_0>0,\quad\forall\ n\in \mathbb{N}\qquad\text{and}\qquad \|w_n\|_{L^\infty}\to 0.$$
which is a contradiction.

Remark 1: Note that in the proof of $(1)$ we have used only two properties of the spaces $W^{1,p}$, $L^\infty$ and $L^1$: (i) the inclusion $W^{1,p}\subset L^\infty$ is compact and (ii) the inclusion $L^\infty\subset L^1$ is continuous. Thus, exactly the same argument can be used to prove the

Ehrling's lemma: Let $A$, $B$ and $X$ be Banach spaces such that $A\subset X\subset B$. Assume that
  
  
*
  
*the inclusion $A\subset X$ is compact;
  
*the inclusion $X\subset B$ is continuous.
  
  
  Then, given $\eta>0$, there exists a constant $C_\eta>0$ such that
  $$\|u\|_X\leq\eta\|u\|_A+C_\eta\|u\|_B,\quad\forall\ u\in A.$$

Remark 2: Assertion $(1)$ can also be proved as a consequence of the Interpolation inequality of Gagliardo–Nirenberg.
