Let $I\subseteq \mathbb{R}$ be an open bounded interval. Show that any $u\in W^{1,p}(I)$ satisfies $|u(x)|\leq C \|u\|_{W^{1,p}(I)}$, where $C>0$ is independent of $u$.
I don't have a lot of high powered inequalities to work with, other than Holder or Minkowski (Young's inequality is probably OK).
I know by FTC that $$u(x) = \int_y^x u'(t) \, dt+u(y).$$ We also have the definition of $\|u\|^p_{W^{1,p}}=\|u\|_{L^p}^p+\|u'\|^p_{L^p}$
And finally we also know (by Minkowski) that for any functions in $L^P(I)$ we have $\|f+g\|_p\leq\|f\|_p+\|g\|_p,$ and that $|a+b|^p\leq 2^p(a^p+b^p)$, and thus $\|f+g\|_p^p\leq(\|f\|_p+\|g\|_p)^p\leq2^p(\|f\|_p^p+\|g\|_p^p)$
I feel like these are the right ingredients to prove the needed result, but I'm struggling to see how I should put them together.
Any thoughts would be greatly appreciated. Thanks in advance.
From the fundamental theorem, we get (putting absolute values everywhere and enlarge integration region from $(x,y)$ to $I$) $$ |u(x)| \le \left| \int_y^x u'(t)dt + u(y)\right| \le \int_y^x |u'(t)|dt +| u(y)| \le \int_I |u'|dt + |u(y)|. $$ Let $L$ be the length of the interval. Now integrate with respect to $y$: $$ L|u(x)| \le \int_I L |u'| + |u| dt \le \max(1,L) \int_I |u'| + |u| dt . $$ In that way, the left-hand side contains the evaluation in $x$, where on the right-hand side we only have integrals. This proves the claim for $p=1$. Apply Hoelder inequality to get the estimate for $p>1$.