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 $K>0$. 
 A: 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
$$||u||_\infty \le \epsilon ||u'||_\infty + \left(1 + \frac{1}{2\epsilon}\right) ||u||_1.$$
A: 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...
A: 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 $\varepsilon\gt0$,
$$
\|u\|_{\infty}\leqslant\varepsilon\|u'\|_\infty+\max\left\{2,\frac1{2\varepsilon}\right\}\|u\|_1.
$$
