Let $I\subset \mathbb R$ be a (not necessary bounded) interval and $p>1$. Show that there exists $C>0$ such that for all $u\in W^{1,p}(I),$ $$ \|u\|_\infty\le C\|u\|_{W^{1,p}(I)} $$
Recalling that $$\|u\|^p_{W^{1,p}(I)} = \|u\|^p_{L^{p}(I)}+\|u'\|^p_{L^{p}(I)}$$ I started with the following well known identity for all $u\in W^{1,p}(I),$
$$u(y)-u(x)=\int_x^y u'(t) \, dt $$ This is true only on the real line. But I dont know how to proceed.
There might be a more elementary way to prove this, but the tools developed by Adams and Fournier in Sobolev Spaces make for a very easy proof (which can be used to verify that the proposition is correct, even in the case that the domain is unbounded). Note that for an interval $U\subset \mathbb{R}$, $U$ satisfies the strong local Lipschitz property (see page 83 of Sobolev Spaces). Therefore, for $p > 1$, we have the imbedding $W^{1,p}(U)\to C^{0,\gamma}(\overline{U})$ for $\gamma = 1-\frac{1}{p}$, i.e. for $u\in W^{1,p}(U)$, there is a version $u^*\in C^{0,\gamma}(\overline{U})$ of $u$ such that $\|u^*-u\|_{W^{1,p}(U)} = 0$ and $$\|u^*\|_{C^{0,\gamma}(\overline{U})}\lesssim_{U,p} \|u\|_{W^{1,p}(U)}$$ Of course, $$\|u^*\|_{C^{0,\gamma}(\overline{U})} = \|u^*\|_{C^0(\overline{U})}+\sup_{x,y\in \overline{U}} \frac{\lvert u^*(x)-u^*(y)\rvert}{\lvert x-y\rvert^{\gamma}}\geq \|u^*\|_{C^0(\overline{U})}$$ and $\|u^*\|_{C^0(\overline{U})} = \|u^*\|_{C^0(U)}$, so $$\|u^*\|_{C^0(U)}\lesssim_{U,p} \|u\|_{W^{1,p}(U)}$$ As $u^*$ is continuous, we have $\|u^*\|_{C^0(U)} = \|u^*\|_{L^{\infty}(U)}$, and as $u^* = u$ almost everywhere, we have $\|u^*\|_{L^{\infty}(U)} = \|u\|_{L^{\infty}(U)}$. Therefore, $$\|u\|_{L^{\infty}(U)}\lesssim_{U,p} \|u\|_{W^{1,p}(U)}$$
You can just follow the proof of Morrey. Taking a peek at Evans, we have
$$u(x) = ⨏_{B(x,1)} u(y) + (u(x)-u(y)) \,dy ≤ ⨏_{B(x,1)} |u(y)|\,dy + ⨏_{B(x,1)} |u(x)-u(y)| \,dy$$ Of course, we are in dimension 1 so $⨏_{B(x,1)}= \frac{1}{2}∫_{x-1}^{x+1}$. The first term is bounded by $ ‖u‖_{L^p}2^{-1/p}<∞ $. For the second, by your given expression,
$$|u(y) - u(x)| = \left| ∫_x^y u'(t) \, dt \right| ≤ ‖u'‖_{L^p} |x-y|^{1-1/p}$$ Hence, $$\text{2nd term} ≤‖u'‖_{L^p}⨏_{B(x,1)} |y-x|^{1-1/p} \, dy = \frac{1}{2}‖u'‖_{L^p} ∫_{-1}^1 |r|^{1-1/p} \, dr = C_p ‖u'‖_{L^p} < \infty $$ So we have a pointwise bound on $u(x)$ uniformly in $x$, from which $‖u‖_{L^∞} < ∞$.
Let's first consider $I$ where the Lebesgue measure $\mu(I)<\infty$. We can use the following nice result (which can be derived from Holder's inequality):
If $I$ has finite (Lebesgue) measure, and we have $1<p\leq q\leq \infty$, $||f||_{L^p(I)}\leq\mu(I)^{1/p-1/q}||f||_{L^q(I)}$, so we have $L^q(I)\subset L^p(I)$. In particular, we have $L^{\infty}(I)\subset L^p(I)$ for $1\leq p<\infty$. Now note that $W^{1,p}(I)\subset L^p(I)$, so in particular, we have that $L^{\infty}(I)\subset W^{1,p}(I)$ for $1\leq p<\infty$.
The desired constant $C$ can just be set to $C=\mu(I)^{1/p}$ (since $q=\infty$ and $1/q=0$). This argument can be extended for Sobolev spaces with $k$ weak derivatives.
Now if $\mu(I)=\infty$, there is no containment among $L^p$ spaces. For now, let's consider $I=(0,\infty)$, which certainly has infinite measure. Consider $f(x)=1/\sqrt x$, $f\in L^1(I)$, but $f\notin L^2(I)$, likewise if we have $g(x)=x^{-1/3}$, $g\in L^1(I)$, but $g\notin L^3(I)$. In general, we can consider $x^{-1/p}$, which will be in $L^1(I)$ but not $L^p(I)$. Also, note that $x^{-1/p}$ is not bounded a.e. so $x^{-1/p}\notin L^{\infty}$, and so there cannot be such a constant $C$.
