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Here is my two-cent insight based on the physicists obsession with units. Suppose you are interested in a Sobolev $(k,p)$-norm on an $N$-dimensional Riemann manifold $(M,g)$,

$$\Vert u\Vert_{k,p} =\left(\int_M |\nabla^k u|^p dV_g\right)^{1/p}.$$

We start by observing that $\nabla^k u$ is measured in $meters^{-k}$ (think that $\nabla^k u=d^ku/dx^k$ and $dx$ is measured in $meters$ while $u$ is a dimensionless quantity, i.e., $0$-density). Hence $|\nabla^k u|^p$ is measured in $meters^{-kp}$. The volume density $dV_g$ is measured in $meters^N$ so that $\int_M |\nabla^k u|^p dV_g$ is measured in $meters^{N-kp}$. This shows that $\Vert u\Vert_{k,p}$ is measured in $meters^{N/p-k}$. Set

$$w_N(k,p):= N/p-k.$$

Then on compact manifolds we have continuous embedding

$$W^{k_1,p_1}\subset W^{k_2,p_2} \Longleftrightarrow -k_1 \leq -k_2,\;\;w_N(k_1,p_1)\leq w_N(k_2,p_2).$$

The embedding is compact if the inequalities in the right-hand-side are strict. To inlcude include the Holder spaces in this story think

$$C^{k,\alpha}= W^{k+\alpha, \infty}.$$

2 deleted 4 characters in body

Here is my two-cent insight based on the physicists obsession with units. Suppose you are interested in a Sobolev norm $(k,p)$ norm (k,p)$-norm on an$N$-dimensional Riemann manifold$(M,g)$(M,g)$,

$$\Vert u\Vert_{k,p} =\left(\int_M |\nabla^k u|^p dV_g\right)^{1/p}.$$

We start by observing that $\nabla^k u$ is measured in $meters^{-k}$ (think that $\nabla^k u=d^ku/dx^k$ and $dx$ is measured in $meters$ while $u$ is a dimensionless quantity, i.e., $0$-density). Hence $|\nabla^k u|^p$ is measured in $meters^{-kp}$. The volume density $dV_g$ is measured in $meters^N$ so that $\int_M |\nabla^k u|^p dV_g$ is measured in $meters^{N-kp}$. This shows that $\Vert u\Vert_{k,p}$ is measured in $meters^{N/p-k}$. Set

$$w_N(k,p):= N/p-k.$$

Then on compact manifolds we have continuous embedding

$$W^{k_1,p_1}\subset W^{k_2,p_2} \Longleftrightarrow -k_1 \leq -k_2,\;\;w_N(k_1,p_1)\leq w_N(k_2,p_2).$$

The embedding is compact if the inequalities in the right-hand-side are strict. To inlcude the Holder spaces in this story think

$$C^{k,\alpha}= W^{k+\alpha, \infty}.$$

1

Here is my two-cent insight based on the physicists obsession with units. Suppose you are interested in a Sobolev norm $(k,p)$ norm on an $N$-dimensional Riemann manifold $(M,g)$

$$\Vert u\Vert_{k,p} =\left(\int_M |\nabla^k u|^p dV_g\right)^{1/p}.$$

We start by observing that $\nabla^k u$ is measured in $meters^{-k}$ (think that $\nabla^k u=d^ku/dx^k$ and $dx$ is measured in $meters$ while $u$ is a dimensionless quantity, i.e., $0$-density). Hence $|\nabla^k u|^p$ is measured in $meters^{-kp}$. The volume density $dV_g$ is measured in $meters^N$ so that $\int_M |\nabla^k u|^p dV_g$ is measured in $meters^{N-kp}$. This shows that $\Vert u\Vert_{k,p}$ is measured in $meters^{N/p-k}$. Set

$$w_N(k,p):= N/p-k.$$

Then on compact manifolds we have continuous embedding

$$W^{k_1,p_1}\subset W^{k_2,p_2} \Longleftrightarrow -k_1 \leq -k_2,\;\;w_N(k_1,p_1)\leq w_N(k_2,p_2).$$

The embedding is compact if the inequalities in the right-hand-side are strict. To inlcude the Holder spaces in this story think

$$C^{k,\alpha}= W^{k+\alpha, \infty}.$$