# A Cauchy with $\varepsilon$- type inequality for $C^1$ functions

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? –  Martin Argerami Mar 9 '12 at 1:19
Typo, $u'$ is continuous. –  checkmath Mar 9 '12 at 1:37
What is «a Cauchy»? –  Mariano Suárez-Alvarez Mar 9 '12 at 1:42
Hey Downvoters could you just explain your point? –  checkmath Mar 9 '12 at 2:55
Perhaps you should explain the context of this problem a bit more. –  Arthur Fischer Mar 9 '12 at 15:47

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...

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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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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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