We first prove the one-dimensional case, which illustrates the key ideas. The result is: For $u \in C_c^1(\mathbb{R})$: $\|u\|_{L^\infty(\mathbb{R})} \leq \frac{1}{2}\|u'\|_{L^1(\mathbb{R})}$

Step 1: Fundamental Theorem of Calculus

For any $x \in \mathbb{R}$: $u(x) = \int_{-\infty}^x u'(t)\,dt$ since $u$ has compact support (so $u(-\infty) = 0$).

Similarly: $-u(x) = \int_x^\infty u'(t)\,dt$

Step 2: Average the Two Representations

Adding and dividing by 2: $|u(x)| = \frac{1}{2}\left|\int_{-\infty}^x u'(t)\,dt - \int_x^\infty u'(t)\,dt\right| \leq \frac{1}{2}\int_{-\infty}^\infty |u'(t)|\,dt$

Therefore: $\|u\|_{L^\infty} \leq \frac{1}{2}\|u'\|_{L^1}$

For $u \in C_c^1(\mathbb{R}^n)$ with $1 \leq p < n$: $\|u\|_{L^{p^*}} \leq C(n,p)\|\nabla u\|_{L^p}$ where $p^* = \frac{np}{n-p}$.

Step 1: Representation Formula

For $x = (x_1, \ldots, x_n)$, integrating along a ray: $|u(x)| = \left|\int_{-\infty}^{x_1} \frac{\partial u}{\partial x_1}(t, x_2, \ldots, x_n)\,dt\right| \leq \int_{-\infty}^\infty \left|\frac{\partial u}{\partial x_1}\right|\,dt$

Step 2: Apply Hölder's Inequality

$|u(x)| \leq \left(\int_{\mathbb{R}} 1^q\,dt\right)^{1/q}\left(\int_{\mathbb{R}} \left|\frac{\partial u}{\partial x_1}\right|^p\,dt\right)^{1/p}$ where $1/p + 1/q = 1$ and $q = p/(p-1)$.

If $p > 1$, the first integral diverges. We need a more sophisticated approach.

Step 3: Iteration (Sketch)

Use representation formulas in each coordinate direction and combine carefully. The full proof requires:

• Estimating $|u|^{p^*}$ by iterating over dimensions
• Using Young's convolution inequality
• Obtaining the sharp constant via Riesz potentials

The result is: $\|u\|_{L^{np/(n-p)}} \leq C\|\nabla u\|_{L^p}$

Step 4: Extension to Sobolev Spaces

By density of $C_c^\infty$ in $W^{1,p}$ and continuity of norms, the inequality extends to all $u \in W^{1,p}(\mathbb{R}^n)$.