Case $p = 1$: For $u \in W^{1,1}(a,b)$, the weak derivative $u'$ is in $L^1$. By the fundamental theorem of calculus for Sobolev functions, $u(x) = u(c) + \int_c^x u'(t) \, dt$ for some $c \in (a,b)$ and almost every $x \in (a,b)$.

For any $x, y \in [a,b]$: $|u(x) - u(y)| \leq \int_y^x |u'(t)| \, dt \leq \|u'\|_{L^1}$

Taking $y$ such that $|u(y)| \leq \|u\|_{L^1}/(b-a)$ (mean value), we get $|u(x)| \leq |u(y)| + |u(x) - u(y)| \leq \frac{\|u\|_{L^1}}{b-a} + \|u'\|_{L^1} \leq C\|u\|_{W^{1,1}}$

Case $p > 1$: For $x, y \in [a,b]$ with $x < y$, Hölder's inequality gives: $|u(x) - u(y)| = \left|\int_x^y u'(t) \, dt\right| \leq \|u'\|_{L^p} |y - x|^{1/p'}$

where $1/p + 1/p' = 1$. Thus $u$ is Hölder continuous with exponent $\alpha = 1 - 1/p = 1/p'$.

For the embedding constant: $\|u\|_{C^{0,\alpha}} = \sup_x |u(x)| + \sup_{x \neq y} \frac{|u(x) - u(y)|}{|x - y|^\alpha} \leq C\|u\|_{W^{1,p}}$

Step 1: For $u \in C_c^1(\mathbb{R}^n)$, write $|u(x)|^{p^*} = p^* \int_{-\infty}^{x_1} \frac{\partial}{\partial t} |u(t, x_2, \ldots, x_n)|^{p^*} \, dt$

Step 2: Apply Hölder's inequality with careful exponents: $|u(x)|^{p^*} \leq C \int_{-\infty}^{x_1} |\nabla u| |u|^{p^*-1} \, dt$

Step 3: Integrate over $\mathbb{R}^n$ and use Hölder again to separate $u$ and $\nabla u$ terms

Step 4: Solve the resulting inequality to get $\|u\|_{L^{p^*}}^{p^*} \leq C \|\nabla u\|_{L^p}^p \|u\|_{L^{p^*}}^{p^*-p}$

Step 5: Rearrange to obtain $\|u\|_{L^{p^*}} \leq C \|\nabla u\|_{L^p}$

Step 6: Extend to $W^{1,p}$ by density of $C_c^1$