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 $$\lVert u\rVert_{\infty}\leq \frac{\varepsilon}{1-\varepsilon}\lVert u'\rVert_{\infty}+\frac{K_{\varepsilon}}{1-\varepsilon}\lVert u\rVert_{L^1}.$$

| cite | improve this answer | |
  • 1
    $\begingroup$ 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
  • 1
    $\begingroup$ @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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.