On a Sobolev embedding: every $u\in W^{1,p}(I)$ is Bounded Let $I\subset \mathbb R$ be a (not necessary bounded) interval and $p&gt;1$. Show that there exists $C&gt;0$ 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 $p &gt; 1$, 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, $$\|u\|_{L^{\infty}(U)}\lesssim_{U,p} \|u\|_{W^{1,p}(U)}$$
Let's first consider $I$ where the Lebesgue measure $\mu(I)&lt;\infty$. We can use the following nice result (which can be derived from Holder's inequality):

If $I$ has finite (Lebesgue) measure, and we have $1&lt;p\leq q\leq \infty$, $||f||_{L^p(I)}\leq\mu(I)^{1/p-1/q}||f||_{L^q(I)}$, so we have $L^q(I)\subset L^p(I)$. In particular, we have $L^{\infty}(I)\subset L^p(I)$ for $1\leq p&lt;\infty$. 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 $1\leq p&lt;\infty$. 

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 $C$.