- Define the space $D(ℝ^n)$ of test functions on $ℝ^n$. What does it mean to say that $u$ is a distribution on $ℝ^n$? What does it mean to say that $u$ is a regular distribution on $ℝ^n$?
- Define for a distribution $u$ on $ℝ^n$ its distributional partial derivatives and explain why they are consistent with the usual partial derivatives for $\mathrm{C}^1$ functions.
- State the definition of the Sobolev space $\mathrm{W}^{1,1}(ℝ^n)$ and its norm.
- Let
\[
f(x)=\begin{cases}
0 & x>1, \\
1-x & 0<x ⩽ 1, \\
1+x & -1<x ⩽ 0, \\
0 & x ⩽-1 ;
\end{cases} \text{ and } g(x)=\begin{cases}
0 & x>1, \\
1-x & 0<x ⩽ 1, \\
2+x & -2<x ⩽ 0, \\
0 & x ⩽-2 .
\end{cases}
\]
Calculate the distributional derivatives $f'$ and $g'$ and decide about membership of $\mathrm{W}^{1,1}(ℝ)$.
- $𝒟(ℝ^n)$ is the vector space of $\mathrm{C}^{∞}$ functions $ϕ: ℝ^n → ℂ$ whose support, $\operatorname{supp}(ϕ)$, defined as the closure in $ℝ^n$ of the set $\{x ∈ ℝ^n:ϕ(x) ≠ 0\}$, is compact. It is equipped with a notion of convergence: $ϕ_j → ϕ$ in $𝒟(ℝ^n)$ means that $ϕ_j, ϕ ∈ 𝒟(ℝ^n)$, and there exists a compact subset $K$ of $ℝ^n$ such that $\operatorname{supp}(ϕ_j), \operatorname{supp}(ϕ) ⊆ K$ and $∂^α ϕ_j → ∂^α ϕ$ uniformly on $ℝ^n$ for each multi-index $α ∈ ℕ_0^n$.
$u ∈ 𝒟'(ℝ^n)$ if $u: 𝒟(ℝ^n) → ℂ$ is a linear and $𝒟$-continuous functional. The latter means that $⟨u, ϕ_j⟩ →⟨u, ϕ⟩$ whenever $ϕ_j → ϕ$ in $𝒟(ℝ^n)$.
$u$ is a regular distribution on $ℝ^n$ if $u ∈ \mathrm{L}_\text{loc}^1(ℝ^n)$, that is, $u$ is represented by a locally integrable function again denoted $u:⟨u, ϕ⟩=∫_{ℝ^n} u ϕ$.
- The distributional partial derivatives $∂_j u$ are defined by the rule $⟨∂_j u, ϕ⟩=-⟨u, ∂_j ϕ⟩, ϕ ∈ 𝒟(ℝ^n)$, where $∂_j ϕ$ denotes the usual partial derivative. Hereby $∂_j u$ is a well-defined distribution on $ℝ^n$ since the map $ϕ ↦-∂_j ϕ$ is linear and $𝒟$-continuous.
This definition is consistent with the usual partial derivatives when $u$ is represented by a $\mathrm{C}^1$ function (again denoted $u$): If $ϕ ∈ 𝒟(ℝ^n)$ we get by use of Fubini's theorem and integration by parts (writing $∂_{x_j}$ for usual partial derivative)
\begin{aligned}
⟨u,-∂_{x_j} ϕ⟩ & =-∫_{ℝ^n} u(x) ∂_{x_j} ϕ(x) \mathrm{d} x=-∫_{ℝ^{n-1}} ∫_{-∞}^{∞} u(x_j, y) ∂_{x_j} ϕ(x_j, y) \mathrm{d} x_j \mathrm{~d} y \\
& =-∫_{ℝ^{n-1}}([u(x_j, y) ϕ(x_j, y)]_{x_j →-∞}^{x_j → ∞}-∫_{-∞}^{∞} ∂_{x_j} u\ ϕ \mathrm{d} x_j) \mathrm{d} y \\
& =∫_{ℝ^n} ∂_{x_j} u\ ϕ \mathrm{d} x .
\end{aligned}
-
\[
\mathrm{W}^{1,1}(ℝ^n)=\{u ∈ 𝒟'(ℝ^n): ∂^α u ∈ \mathrm{L}^1(ℝ^n) \text{ for }|α| ≤ 1\}
\]
and
\[
\|u\|_{\mathrm{W}^{1,1}}=\|u\|_1+\sum_{j=1}^n\|∂_j u\|_1 .
\]
- The functions $f, g$ are clearly integrable on $ℝ$ so are in particular regular
distributions on $ℝ$. We have $f'=𝟏_{(-1,0)}-𝟏_{(0,1)}$ since for $ϕ ∈ 𝒟(ℝ)$:
\[
\begin{aligned}
⟨f', ϕ⟩ & =-⟨f, ϕ'⟩=-∫_{ℝ} f ϕ' \mathrm{d} x \\
& =-∫_{-1}^0(x+1) ϕ'(x) \mathrm{d} x-∫_0^1(1-x) ϕ'(x) \mathrm{d} x \\
& \stackrel{\text{parts}}{=}-([(x+1) ϕ(x)]_{-1}^0-∫_{-1}^0 ϕ \mathrm{d} x+[(1-x) ϕ(x)]_0^1-∫_0^1(-1) ϕ \mathrm d x) \\
& =∫_{-1}^0 ϕ \mathrm{d} x-∫_0^1 ϕ \mathrm{d} x=∫_{ℝ}(𝟏_{(-1,0)}-𝟏_{(0,1)}) ϕ \mathrm{d} x
\end{aligned}
\]
as required. Since $f, f' ∈ \mathrm{L}^1(ℝ)$ we have $f ∈ \mathrm{W}^{1,1}(ℝ)$.
We have $g'=𝟏_{(-2,0)}-δ_0-𝟏_{(0,1)}$ since for $ϕ ∈ 𝒟(ℝ)$:
\begin{aligned}
⟨g', ϕ⟩ & =-∫_{ℝ} g ϕ' \mathrm{d} x \\
&= -∫_{-2}^0(x+2) ϕ' \mathrm{d} x-∫_0^1(1-x) ϕ' \mathrm{d} x \\
& \stackrel{\text{parts}}{=}-[(x+2) ϕ]_{-2}^0+∫_{-2}^0 ϕ \mathrm{d} x-[(1-x) ϕ]_0^1+∫_0^1(-1) ϕ \mathrm{d} x \\
&= -ϕ(0)+∫_{ℝ}(𝟏_{(-2,0)}-𝟏_{(0,1)}) ϕ \mathrm{d} x
\end{aligned}
as required. Since $g' ∉ \mathrm{L}^1(ℝ)$ it follows that $g ∉ \mathrm{W}^{1,1}(ℝ)$.
Prove that $\mathrm{W}^{1,1}(ℝ^n)$ is complete.
[You may assume that $\mathrm{L}^1(ℝ^n)$ is complete.]
Let $(u_j)$ be a Cauchy sequence in $\mathrm{W}^{1,1}(ℝ^n)$: since
\[
\|u_i-u_j\|_{\mathrm{W}^{1,1}}=\|u_i-u_j\|_1+\sum_{k=1}^n\|∂_k u_i-∂_k u_j\|_1
\]
it follows that the sequences $(u_j),(∂_1 u_j), …,(∂_n u_j)$ are Cauchy in $\mathrm{L}^1(ℝ^n)$. By completeness of $\mathrm{L}^1(ℝ^n)$ we find $u, v_1, …, v_n ∈ \mathrm{L}^1(ℝ^n)$ such that $\|u-u_j\|_1 → 0$ and $\|∂_k u_j-v_k\|_1 → 0$ as $j → ∞$ for each $k ∈\{1, …, n\}$. Let $ϕ ∈ 𝒟(ℝ^n)$. Then
\begin{aligned}
&⟨∂_k u_j, ϕ⟩=∫_{ℝ^n} ∂_k u_j\ ϕ \mathrm{d} x → ∫_{ℝ^n} v_k ϕ \mathrm{d} x, \\
&⟨∂_k u_j, ϕ⟩=-⟨u_j, ∂_k ϕ⟩=-∫_{ℝ^n} u_j ∂_k ϕ \mathrm{d} x →-∫_{ℝ^n} u ∂_k ϕ \mathrm{d} x=⟨∂_k u, ϕ⟩
\end{aligned}
as $j → ∞$. But then $⟨∂_k u, ϕ⟩=∫_{ℝ^n} v_k ϕ \mathrm{d} x$, and since $ϕ$ was an arbitrary test functions, $∂_k u=v_k ∈ \mathrm{L}^1(ℝ^n)$. Thus $u ∈ \mathrm{W}^{1,1}(ℝ^n)$ and since clearly $\|u_j-u\|_{\mathrm{W}^{1,1}} → 0$ we are done.
Give an example of a distribution $u ∈ \mathrm{W}^{1,1}(ℝ^2)$ that does not have a continuous representative.
Put $χ=ρ *𝟏_{B_2(0)}$. Then $χ ∈ 𝒟(ℝ^2)$ and $𝟏_{B_1(0)} ≤ χ ≤𝟏_{B_3(0)}$. Define for an $α ∈(0,1)$ the function $u(x, y)=(x^2+y^2)^{-\frac{α}{2}} χ(x, y),(x, y) ∈ℝ^2$. Note that $u$ is continuous on $ℝ^2 ∖\{0\}$ hence is measurable, hence as $∫_{ℝ^2}|u(x, y)| \mathrm{d}(x, y) ≤2π ∫_0^3 r^{1-α} \mathrm{d} r=2π 3^{2-α} /(2-α)<+∞$ we have $u ∈\mathrm{L}^1(ℝ^n)$. The distribution $u$ does not have a continuous representative since if $f ∈ \mathrm{C}(ℝ^n)$ and $m=\max _{\overline{B_3(0)}} f(x, y)$, then for $r=|(x, y)|<\min \{1, m^{-1 / α}\}$ we have $u(x, y)=r^{-α}>m ≥ f(x, y)$ and consequently $u$ and $f$ cannot agree almost everywhere.
The function $u$ is symmetric in $x, y$ so it suffices to prove that $∂_x u ∈\mathrm{L}^1(ℝ^2)$. First we calculate the usual partial derivative of $u$ on $ℝ^2 ∖\{0\}$: $u_x=-α x r^{-α-2} χ+r^{-α} χ_x$. The function is continuous on $ℝ^2 ∖\{0\}$ hence measurable and (since $|x| ≤ r$)
\[
∫_{ℝ^2}|u_x| \mathrm{d}(x, y) ≤2π ∫_0^3(α r^{-α}+r^{1-α} \max |χ_x|) \mathrm{d} r<+∞
\]
since $α<1$ it follows by Tonelli's theorem that $u_x ∈ \mathrm{L}^1(ℝ^2)$. Next we must check that $u_x$ equals the distributional partial derivative $∂_x u$. For each $y ≠ 0$, $u_x$ is continuous, we calculate for a test function $ϕ ∈ 𝒟(ℝ^2)$ by integration by parts:
\[
∫_{-∞}^{∞} u ϕ_x \mathrm{~d} x=-∫_{-∞}^{∞} u_x ϕ \mathrm{d} x
\]
and since $ℝ ×\{0\}$ is a null set we get by use of Fubini's theorem that
\[
⟨u, ϕ_x⟩=-∫_{ℝ} ∫_{ℝ} u_x ϕ \mathrm{d} x \mathrm{~d} y=-∫_{ℝ^2} u_x ϕ \mathrm{d}(x, y) .
\]
Thus $∂_x u=u_x ∈ \mathrm{L}^1(ℝ^2)$ and therefore $u ∈ \mathrm{W}^{1,1}(ℝ^2)$.
Conclude that the convergence $u_j → u$ in $\mathrm{W}^{1,1}(ℝ^2)$ does not entail that $\|u_j-u\|_{∞} → 0$
Put $u_j=ρ_{1 / j} * u$. Then $u_j ∈ \mathrm{C}^{∞}(ℝ^2)$ and since $∂_s u_j=ρ_{1 / j} *∂_s u$ it follows by Fubini that $\|u_j\|_1 ≤\|u\|_1,\|∂_s u_j\|_1 ≤\|∂_s u\|_1$ (for $s ∈\{x, y\})$, so $u_j ∈ \mathrm{W}^{1,1}(ℝ^2)$ too and $\|u_j-u\|_{\mathrm{W}^{1,1}} → 0$. But $u_j$ is in particular continuous, so if $\|u_j-u\|_{∞} → 0$, then the triangle inequality yields $\| u_j-u_k\|_{∞} ≤\| u_j-u\|_{∞}+\| u-u_k \|_{∞} → 0$, that is, $(u_j)$ is Cauchy in $\mathrm{L}^{∞}(ℝ^2)$ and so is a uniform Cauchy sequence. It then converges uniformly and so $\lim _{j → ∞} u_j$ is a continuous function. By Fatou's lemma $\|\lim _{j → ∞} u_j-u\|_1 ≤\lim _{j → ∞}\|u_j-u\|_1=0$ and so $\lim _{j → ∞} u_j$ is a continuous representative for $u$, a contradiction concluding the proof.