Let $X,Y$ and $Z$ be three Banach spaces with norms $\|\cdot\|_X$, $\|\cdot\|_Y$,$\|\cdot\|_Z$. Assume that $X\subset Y$ with compact "injection" and that $Y\subset Z$ with continuous injection. Then

$$\forall\epsilon>0, \exists C_{\epsilon}\geq0 $$

Satisfying $$\|u\|_Y\leq \epsilon \|u\|_X+C_{\epsilon}\|u\|_Z \ \ \forall u \in X.$$

My question are

I) Where can I find a proof of this result?

II) As a consequence of that how to prove

$$\max_{[0,1]}|u|\leq \epsilon\max_{[0,1]}|u'|+C_{\epsilon}\|u\|_{L^1{[0,1]}} \forall \in C^1({[0,1]})?$$

  • $\begingroup$ You can find this in Brezis, Functional Analysis, Sobolev Spaces and Partial Differential Equations, 2011, Exercise 6.12, pag. 173. I think there are also the answers for the exercise in the back of the book. $\endgroup$ – Beni Bogosel Mar 17 '12 at 19:17

If the result were not true, we would be able to find $\varepsilon_0>0$ and a sequence $\{u_n\}\subset X$ such that $\lVert u_n\rVert_Y=1$ and $1\geq \varepsilon_0\lVert u_n\rVert_X+n\lVert u_n\rVert_Z$. Since the sequence $\{u_n\}$ is bounded in $X$ and the inclusion $X\subset Y$ is compact, we can find a subsequence denoted $\{v_k\}$ which converges to $v$ in $Y$. Since the inclusion $Y\subset Z$ is continuous, $v_n\to v$ in $Z$. This gives a contradiction since $\lVert v_k\rVert_Z\leq \frac 1k$ so $v=0$ but $\lVert v_k\rVert_Y=1$ so $\lVert v\rVert_Y$.

For the second question, take $X$ the Banach space of continuously differentiable functions endowed with the norm $\lVert u\rVert_X:=\lVert u\rVert_{\infty}+\lVert u'\rVert_{\infty}$, $Y$ the space of continuous functions on $[0,1]$ endowed with the norm $\lVert u\rVert_Y:=\lVert u\rVert_{\infty}$ and $Z=L^1[0,1]$ with the natural norm. For a fixed $\varepsilon>0$, we get a constant $K_{\varepsilon}$ such that for all $u\in X$: $\lVert u\rVert_{\infty}\leq \varepsilon(\lVert u\rVert_{\infty}+\lVert u'\rVert_{\infty})+K_{\varepsilon}\lVert u\rVert_{L^1}$ so fo $\varepsilon< 1$ $$(1-\varepsilon)\lVert u\rVert_{\infty}\leq \varepsilon\lVert u'\rVert_{\infty}+K_\varepsilon\lVert u\rVert_{L^1},$$ and so $$\lVert u\rVert_{\infty}\leq \frac{\varepsilon}{1-\varepsilon}\lVert u'\rVert_{\infty}+\frac{K_{\varepsilon}}{1-\varepsilon}\lVert u\rVert_{L^1}.$$

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    $\begingroup$ For a reference (containing the same proof of the abstract result), see e.g. Lemma 1.1, page 106 of Showalter's Monotone operators in Banach spaces and nonlinear partial differential equations. $\endgroup$ – t.b. Mar 9 '12 at 17:58
  • $\begingroup$ @t.b. Thanks for the reference, I didn't know this book. For the last part, the inconvenient of this method is that we don't know the constant $K_{\varepsilon}$, whereas it's possible to determine it with a direct approach. $\endgroup$ – Davide Giraudo Mar 9 '12 at 18:01
  • $\begingroup$ Why does it follow that $\| u_n\|_X$ is bounded? we have $\| \cdot \|_Y \leq C\| \cdot \|_X$ not the other way around. $\endgroup$ – Jose27 Mar 9 '12 at 21:10
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    $\begingroup$ @Jose27 We have $\lVert u_n\rVert_X\leq \frac 1{\varepsilon_0}$. $\endgroup$ – Davide Giraudo Mar 9 '12 at 21:12
  • $\begingroup$ Of course, thank you. $\endgroup$ – Jose27 Mar 9 '12 at 21:36

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