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Let $1 \le p < \infty$. For all $\epsilon > 0$, does there exist $C = C(\epsilon, q)$ such that$$\|u\|_{L^p(0, 1)} \le \epsilon \|u'\|_{L^1(0, 1)} + C\|u\|_{L^1(0, 1)} \text{ for all }u \in W^{1, 1}(0, 1)?$$

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  • $\begingroup$ This is actually Exercise 8.5, 3. in Brezis. $\endgroup$ Feb 17, 2016 at 20:53

3 Answers 3

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Of course.

For $u\in C^1$ denote $v=u-\int_0^1 u$, then $v(x_0)=0$ for some $x_0\in (0,1)$ and for all $x\in (0,1)$ we have $$|v(x)|=|v(x)-v(x_0)|=\left|\int_{x_0}^x v'(t)dt\right|\leqslant \|v'\|_{L^1(0,1)}=\|u'\|_{L^1(0,1)},$$ thus $M:=\|v\|_{L^{\infty}(0,1)}\leqslant \|u'\|_{L^1(0,1)}$. This extends to $u\in W^{1,1}$ by continuity. We have $\|u\|_p\leqslant \|v\|_p+\|u\|_1$. Next, $$ \|v\|_p^p=\int |v(x)|^p dx\leqslant M^{p-1} \int |v(x)| dx= (\varepsilon^{-1} \|v\|_1^p)^{1/p}\cdot (M^{p} \varepsilon^{1/(p-1)})^{1-1/p}\leqslant \varepsilon^{-1} \|v\|_1^p+ M^{p} \varepsilon^{1/(p-1)}, $$ and result follows from this inequality and $\|v\|_1\leqslant 2\|u\|_1$.

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Yes, this is the Sobolev injection $W^{1, 1}(]0, 1[) \to C^0([0, 1])$ (see e.g. Brezis' book on functional analysis) $$ \|u\|_{L^\infty} \leq C \left( \|u'\|_{L^1} + \|u\|_{L^1}\right) $$

followed by the $\epsilon$-Young inequality:

\begin{align*} \|u\|_{L^p} & \leq \left(\|u\|_{L^1} \|u\|_{L^\infty}^{p-1}\right)^{1/p}\\ & \leq \frac{\|u\|_{L^1}}{2 \epsilon^p} + \frac{\epsilon^{p/(p-1)}}{2} \|u\|_{L^\infty}. \end{align*}

There is just a slight adjustment to make in the definition of $\epsilon$ if you want to get exactly the inequality you want.

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The inequality you ask for is more or less a restatement of the compactness of an appropriate embedding: Let $p\le q\le r$ be three norms on a vector space such that the ball $B_r(1)=\lbrace x\in X. r(x)\le 1\rbrace$ is relatively compact in $(X,q)$. Then, for every $\varepsilon>0$ there is $C>0$ such that $$ q(x) \le \varepsilon r(x) + C p(x). $$

Indeed, the topologies induced by $p$ and $q$ on $B_r(1)$ coincide because a (relatively) compact space does not admit strictly coarser Hausdorff topologies. For $\varepsilon >0$ there is thus $\delta>0$ such that $$ B_p(\delta)\cap B_r(1) \subseteq B_q(\varepsilon). $$ For $x\in X$ with $r(x)<\infty$ and $t=p(x)/\delta +r(x)$ you have $\frac 1t x \in B_p(\delta)\cap B_r(1)$ and hence $q(x)\le t\varepsilon$ which proves the inequality for $C=\varepsilon/\delta$.

In your case, $r$ is the $W^{1,1}$-norm, $p$ the $L^1$-norm, and $q$ is the $L^p$-norm.

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