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The inequality is the following:

For any $\varepsilon >0$ there exists $C_\varepsilon >0$ s.t.:

$$\lVert u\rVert_{\infty}\leq \varepsilon\ \lVert u^\prime \rVert_{\infty}+C_{\varepsilon}\ \lVert u\rVert_1$$

for any $u\in C^1([0,1])$.

Does anyone know the trick to obtain this inequality? I am pretty sure it is just a little trick.

The constant $C_\varepsilon$ should be something like $\frac{K}{\varepsilon}$ with $K>0$.

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    What is $u'$ here?2012-03-09
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    Typo, $u'$ is continuous.2012-03-09
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    What is «a Cauchy»?2012-03-09
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    Hmm. Bound the max of a continuous function with the max of the derivative and... $C_\epsilon$ (which is something?) times the integral. Hmm.2012-03-09
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    $C_{\epsilon} $ only depends of $\epsilon$ it will be probably $2012-03-09
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    Hey Downvoters could you just explain your point?2012-03-09
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    Perhaps you should explain the context of this problem a bit more.2012-03-09
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    That was a question in an test!2012-03-09
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    *Hey accepter could you just explain your point* in accepting an incomplete solution?2012-03-14
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    That answer is not fully but his idea was almost the same! (Pythagoras theorem holds in Any Hilbert Space what Pythagoras never thought about, but we still call it Pythagoras Theorem)2012-03-14
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    Pythagoras notwithstanding, the point of the exercise is to show that the multiple of $\|u'\|_\infty$ may be made as small as one wants. Hence no, the idea of the accepted answer cannot yield a solution.2012-03-18
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    I know that the answer is not complete ok! I was about to choose none but I solved to choose Picciu's answer, however incomplete you cited it then I thought it would be better than choose yours or the other!2012-03-18

3 Answers 3

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As explained by Pacciu, $|u(x)|\leqslant\|u'\|_\infty\,|x-y|+|u(y)|$ for every $x$ and $y$. Assume first that $\varepsilon\leqslant\frac14$ and $x\geqslant2\varepsilon$ and let us integrate this inequality from $y=x-2\varepsilon$ to $y=x$. The result is $$ 2\varepsilon |u(x)|\leqslant\|u'\|_\infty\,\int_{x-2\varepsilon}^{x}|x-y|\mathrm dy+\int_{x-2\varepsilon}^{x}|u(y)|\mathrm dy\leqslant\|u'\|_\infty\,2\varepsilon^2+\|u\|_1. $$ The same inequality holds if $x\leqslant2\varepsilon$, using the integral from $y=x$ to $y=x+2\varepsilon$.

Thus, if $\varepsilon\leqslant\frac14$, $\|u\|_{\infty}\leqslant\varepsilon \|u'\|_\infty+\frac1{2\varepsilon}\|u\|_1$. In particular, $\|u\|_{\infty}\leqslant\tfrac14\|u'\|_\infty+2\|u\|_1$ hence, for every $\varepsilon\geqslant\frac14$, $\|u\|_{\infty}\leqslant\varepsilon \|u'\|_\infty+2\|u\|_1$.

Finally, for every $\varepsilon\gt0$, $$ \|u\|_{\infty}\leqslant\varepsilon\|u'\|_\infty+\max\left\{2,\frac1{2\varepsilon}\right\}\|u\|_1. $$

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Suppose first that there is some $t_0$ with $u(t_0) = 0$. Then by the fundamental theorem of calculus, we have $$u(t)^2 = 2 \int_{t_0}^t u'(t) u(t)\,dt \le 2 ||u'||_{\infty} ||u||_1.$$ Thus $$|u(t)| \le \sqrt{(2 \epsilon ||u'||_\infty)(\frac{1}{\epsilon} ||u||_1)}$$ So taking the supremum over $t$ and using the AM-GM inequality we get $$||u||_\infty \le \epsilon ||u'||_\infty + \frac{1}{2\epsilon} ||u||_1.$$

Now suppose there is no $t_0$ with $u(t_0) = 0$. By the intermediate value theorem either $u>0$ everywhere or $u<0$ everywhere. By replacing $u$ by $-u$ if necessary we assume $u > 0$ everywhere. Let $t_1$ be the point where $u$ attains its minimum. Set $\tilde{u}(t) = u(t) - u(t_1)$. Note that $\tilde{u}' = u'$ and $||\tilde{u}||_1 = ||u||_1 - u(t_1) \le ||u||_1$. Applying the previous case to $\tilde{u}$ we have $$||u||_\infty = ||\tilde{u}||_\infty + u(t_1) \le \epsilon ||u||_\infty + \frac{1}{2\epsilon} ||u||_1 + u(t_1).$$ Finally, by integrating the inequality $u(t_1) \le u(t) = |u(t)|$, we have $u(t_1) \le ||u||_1$. So putting this together gives $$||u||_\infty \le \epsilon ||u'||_\infty + \left(1 + \frac{1}{2\epsilon}\right) ||u||_1.$$

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The following is not a complete answer, but (hopefully) it could be useful.

It is easy to get the inequality without $\varepsilon$.

In fact, let $u\in C^1([0,1])$. Since $u^\prime$ is bounded (by Weierstrass), $u$ is Lipschitz and: $$|u(x)-u(y)|\leq \lVert u^\prime \rVert_\infty\ |x-y|\; ;$$ by reverse triangle inequality one gets: $$|u(x)|\leq \lVert u^\prime \rVert_\infty\ |x-y| + |u(y)|$$ and an integration w.r.t. $y\in [0,1]$ yields: $$|u(x)|\leq \lVert u^\prime \rVert_\infty\ \left( x^2-x+\frac{1}{2}\right) +\lVert u\rVert_1\; .$$ Maximizing both LH and RH sides w.r.t. $x\in [0,1]$ one obtains: $$\tag{1} \lVert u\rVert_\infty \leq \frac{1}{2}\ \lVert u^\prime \rVert_\infty +\lVert u\rVert_1\; .$$

Therefore a comparison of (1) with your guess for $C_\varepsilon$, i.e. $C_\varepsilon = K/\varepsilon$, yields $C_\varepsilon = \frac{1}{2\varepsilon}$.

Coming to the inequality with $\varepsilon$, I have to think about it...