Fréchet–Kolmogorov theorem

In functional analysis, the Fréchet–Kolmogorov theorem (the names of Riesz or Weil are sometimes added as well) gives a necessary and sufficient condition for a set of functions to be relatively compact in an Lp space. It can be thought of as an Lp version of the Arzelà–Ascoli theorem, from which it can be deduced. The theorem is named after Maurice René Fréchet and Andrey Kolmogorov.

Statement
Let $$B$$ be a subset of $$L^p(\mathbb{R}^n)$$ with $$p\in[1,\infty)$$, and let $$\tau_h f$$ denote the translation of $$f$$ by $$h$$, that is, $$\tau_h f(x)=f(x-h) .$$

The subset $$B$$ is relatively compact if and only if the following properties hold:


 * 1) (Equicontinuous) $$\lim_{|h|\to 0}\Vert\tau_h f-f\Vert_{L^p(\mathbb{R}^n)} = 0$$ uniformly on $$B$$.
 * 2) (Equitight) $$\lim_{r\to\infty}\int_{|x|>r}\left|f\right|^p=0$$ uniformly on $$B$$.

The first property can be stated as $$\forall \varepsilon >0 \, \, \exists \delta >0 $$ such that $$\Vert\tau_h f-f\Vert_{L^p(\mathbb{R}^n)} < \varepsilon \, \, \forall f \in B, \forall h$$ with $$|h|<\delta .$$

Usually, the Fréchet–Kolmogorov theorem is formulated with the extra assumption that $$B$$ is bounded (i.e., $$\Vert f\Vert_{L^p(\mathbb{R}^n)}<\infty$$ uniformly on $$B$$). However, it has been shown that equitightness and equicontinuity imply this property.

Special case
For a subset $$B$$ of $$L^p(\Omega)$$, where $$\Omega$$ is a bounded subset of $$\mathbb{R}^n$$, the condition of equitightness is not needed. Hence, a necessary and sufficient condition for $$B$$ to be relatively compact is that the property of equicontinuity holds. However, this property must be interpreted with care as the below example shows.

Existence of solutions of a PDE
Let $$(u_\epsilon)_\epsilon$$ be a sequence of solutions of the viscous Burgers equation posed in $$\mathbb{R}\times(0,T)$$:


 * $$\frac{\partial u}{\partial t} + \frac{1}{2}\frac{\partial u^2}{\partial x} = \epsilon\Delta u, \quad u(x,0) = u_0(x),$$

with $$u_0$$ smooth enough. If the solutions $$(u_\epsilon)_\epsilon$$ enjoy the $$L^1$$-contraction and $$L^\infty$$-bound properties, we will show existence of solutions of the inviscid Burgers equation


 * $$\frac{\partial u}{\partial t} + \frac{1}{2}\frac{\partial u^2}{\partial x} = 0, \quad u(x,0) = u_0(x).$$

The first property can be stated as follows: If $$u,v$$ are solutions of the Burgers equation with $$u_0,v_0$$ as initial data, then


 * $$\int_{\mathbb{R}}|u(x,t)-v(x,t)|dx\leq \int_{\mathbb{R}}|u_0(x)-v_0(x)|dx.$$

The second property simply means that $$\Vert u(\cdot,t)\Vert_{L^\infty(\mathbb{R})}\leq \Vert u_0\Vert_{L^\infty(\mathbb{R})}$$.

Now, let $$K\subset\mathbb{R}\times(0,T)$$ be any compact set, and define


 * $$w_\epsilon(x,t):=u_\epsilon(x,t)\mathbf{1}_K(x,t),$$

where $$\mathbf{1}_K$$ is $$1$$ on the set $$K$$ and 0 otherwise. Automatically, $$B:=\{(w_\epsilon)_\epsilon\}\subset L^1(\mathbb{R}^2)$$ since


 * $$\int_{\mathbb{R}^2}|w_\epsilon(x,t)|dx dt= \int_{\mathbb{R}^2}|u_\epsilon(x,t)\mathbf{1}_K(x,t)|dx dt\leq \Vert u_0\Vert_{L^\infty(\mathbb{R})}|K|<\infty.$$

Equicontinuity is a consequence of the $$L^1$$-contraction since $$u_\epsilon(x-h,t)$$ is a solution of the Burgers equation with $$u_0(x-h)$$ as initial data and since the $$L^\infty$$-bound holds: We have that


 * $$\Vert w_\epsilon(\cdot-h,\cdot-h)-w_\epsilon\Vert_{L^1(\mathbb{R}^2)}\leq \Vert w_\epsilon(\cdot-h,\cdot-h)-w_\epsilon(\cdot,\cdot-h)\Vert_{L^1(\mathbb{R}^2)}+\Vert w_\epsilon(\cdot,\cdot-h)-w_\epsilon\Vert_{L^1(\mathbb{R}^2)}.$$

We continue by considering


 * $$\begin{align}

&\Vert w_\epsilon(\cdot-h,\cdot-h)-w_\epsilon(\cdot,\cdot-h)\Vert_{L^1(\mathbb{R}^2)}\\ &\leq \Vert (u_\epsilon(\cdot-h,\cdot-h)-u_\epsilon(\cdot,\cdot-h))\mathbf{1}_K(\cdot-h,\cdot-h)\Vert_{L^1(\mathbb{R}^2)}+\Vert u_\epsilon(\cdot,\cdot-h)(\mathbf{1}_K(\cdot-h,\cdot-h)-\mathbf{1}_K(\cdot,\cdot-h)\Vert_{L^1(\mathbb{R}^2)}. \end{align} $$

The first term on the right-hand side satisfies


 * $$\Vert (u_\epsilon(\cdot-h,\cdot-h)-u_\epsilon(\cdot,\cdot-h))\mathbf{1}_K(\cdot-h,\cdot-h)\Vert_{L^1(\mathbb{R}^2)}\leq T\Vert u_0(\cdot-h)-u_0\Vert_{L^1(\mathbb{R})}$$

by a change of variable and the $$L^1$$-contraction. The second term satisfies


 * $$\Vert u_\epsilon(\cdot,\cdot-h)(\mathbf{1}_K(\cdot-h,\cdot-h)-\mathbf{1}_K(\cdot,\cdot-h))\Vert_{L^1(\mathbb{R}^2)}\leq \Vert u_0\Vert_{L^\infty(\mathbb{R})}\Vert \mathbf{1}_K(\cdot-h,\cdot)-\mathbf{1}_K\Vert_{L^1(\mathbb{R}^2)}$$

by a change of variable and the $$L^\infty$$-bound. Moreover,


 * $$\Vert w_\epsilon(\cdot,\cdot-h)-w_\epsilon\Vert_{L^1(\mathbb{R}^2)}\leq \Vert (u_\epsilon(\cdot,\cdot-h)-u_\epsilon)\mathbf{1}_K(\cdot,\cdot-h)\Vert_{L^1(\mathbb{R}^2)}+\Vert u_\epsilon(\mathbf{1}_K(\cdot,\cdot-h)-\mathbf{1}_K)\Vert_{L^1(\mathbb{R}^2)}.$$

Both terms can be estimated as before when noticing that the time equicontinuity follows again by the $$L^1$$-contraction. The continuity of the translation mapping in $$L^1$$ then gives equicontinuity uniformly on $$B$$.

Equitightness holds by definition of $$(w_\epsilon)_\epsilon$$ by taking $$r$$ big enough.

Hence, $$B$$ is relatively compact in $$L^1(\mathbb{R}^2)$$, and then there is a convergent subsequence of $$(u_\epsilon)_\epsilon$$ in $$L^1(K)$$. By a covering argument, the last convergence is in $$L_{loc}^1(\mathbb{R}\times(0,T))$$.

To conclude existence, it remains to check that the limit function, as $$\epsilon\to0^+$$, of a subsequence of $$(u_\epsilon)_\epsilon$$ satisfies


 * $$\frac{\partial u}{\partial t} + \frac{1}{2}\frac{\partial u^2}{\partial x} = 0, \quad u(x,0) = u_0(x).$$