Does convergence in $L^1_{text{loc}}(mathbb{R}^n)$ implies convergence in $mathcal{D}'(mathbb{R}^n)$?
Every function $fin L^1_{text{loc}}(mathbb{R}^n)$ defines an element $Lambda_f$ of $mathcal{D}'(mathbb{R}^n)$.
If $f_nto f$ in $L^1_{text{loc}}(mathbb{R}^n)$, does that mean that $Lambda_{f_n}to Lambda_f$ in $mathcal{D}'(mathbb{R}^n)$?
distribution-theory
asked Jan 4 at 12:12
Gabriel RibeiroGabriel Ribeiro
1,408422
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Every function $fin L^1_{text{loc}}(mathbb{R}^n)$ defines an element $Lambda_f$ of $mathcal{D}'(mathbb{R}^n)$.
If $f_nto f$ in $L^1_{text{loc}}(mathbb{R}^n)$, does that mean that $Lambda_{f_n}to Lambda_f$ in $mathcal{D}'(mathbb{R}^n)$?
distribution-theory
asked Jan 4 at 12:12
Gabriel RibeiroGabriel Ribeiro
1,408422
add a comment |
Every function $fin L^1_{text{loc}}(mathbb{R}^n)$ defines an element $Lambda_f$ of $mathcal{D}'(mathbb{R}^n)$.
If $f_nto f$ in $L^1_{text{loc}}(mathbb{R}^n)$, does that mean that $Lambda_{f_n}to Lambda_f$ in $mathcal{D}'(mathbb{R}^n)$?
distribution-theory
asked Jan 4 at 12:12
Gabriel RibeiroGabriel Ribeiro
1,408422
Every function $fin L^1_{text{loc}}(mathbb{R}^n)$ defines an element $Lambda_f$ of $mathcal{D}'(mathbb{R}^n)$.
If $f_nto f$ in $L^1_{text{loc}}(mathbb{R}^n)$, does that mean that $Lambda_{f_n}to Lambda_f$ in $mathcal{D}'(mathbb{R}^n)$?
distribution-theory
distribution-theory
asked Jan 4 at 12:12
Gabriel RibeiroGabriel Ribeiro
1,408422
asked Jan 4 at 12:12
Gabriel RibeiroGabriel Ribeiro
1,408422
asked Jan 4 at 12:12
Gabriel RibeiroGabriel Ribeiro
1,408422
asked Jan 4 at 12:12
Gabriel RibeiroGabriel Ribeiro
1,408422
asked Jan 4 at 12:12
Gabriel RibeiroGabriel Ribeiro
1,408422
1,408422
add a comment |
add a comment |
1 Answer
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Yes. The point is that convergence in $L^1_{mathrm{loc}}(mathbb R^n)$ implies convergence in each $L^1(Omega)$ for every bounded domain $Omega subset mathbb R^n,$ and hence weak $L^1$ convergence in $Omega.$ But $L^1(Omega)^* cong L^{infty}(Omega),$ which contains $C^{infty}_c(Omega).$
Going along these lines, we can check this directly. Suppose $f_n rightarrow f$ in $L^1_{mathrm{loc}}(mathbb R^n)$ and let $varphi in C^{infty}_c(mathbb R^n).$ Pick a bounded domain $Omega subset mathbb R^n$ such that $operatorname{supp} varphi subset Omega$ and note that $f_n rightarrow f$ in $L^1(Omega).$ Then we get,
$$ Lambda_{f_n}(varphi) - Lambda_f(varphi) = int_{mathbb R^n} (f_n - f)varphi ,mathrm{d}x = int_{Omega} (f_n - f)varphi ,mathrm{d}x leq lVert f_n - frVert_{L^1(Omega)} lVert varphi rVert_{L^{infty}_{Omega}} longrightarrow 0 $$
as $n rightarrow infty.$ Since $varphi$ was arbitrary, $Lambda_{f_n} rightarrow Lambda_f$ in $mathcal{D}'(mathbb R^n).$
answered Jan 4 at 12:36
ktoiktoi
2,3331616
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1 Answer
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1 Answer
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Yes. The point is that convergence in $L^1_{mathrm{loc}}(mathbb R^n)$ implies convergence in each $L^1(Omega)$ for every bounded domain $Omega subset mathbb R^n,$ and hence weak $L^1$ convergence in $Omega.$ But $L^1(Omega)^* cong L^{infty}(Omega),$ which contains $C^{infty}_c(Omega).$
Going along these lines, we can check this directly. Suppose $f_n rightarrow f$ in $L^1_{mathrm{loc}}(mathbb R^n)$ and let $varphi in C^{infty}_c(mathbb R^n).$ Pick a bounded domain $Omega subset mathbb R^n$ such that $operatorname{supp} varphi subset Omega$ and note that $f_n rightarrow f$ in $L^1(Omega).$ Then we get,
$$ Lambda_{f_n}(varphi) - Lambda_f(varphi) = int_{mathbb R^n} (f_n - f)varphi ,mathrm{d}x = int_{Omega} (f_n - f)varphi ,mathrm{d}x leq lVert f_n - frVert_{L^1(Omega)} lVert varphi rVert_{L^{infty}_{Omega}} longrightarrow 0 $$
as $n rightarrow infty.$ Since $varphi$ was arbitrary, $Lambda_{f_n} rightarrow Lambda_f$ in $mathcal{D}'(mathbb R^n).$
answered Jan 4 at 12:36
ktoiktoi
2,3331616
add a comment |
Yes. The point is that convergence in $L^1_{mathrm{loc}}(mathbb R^n)$ implies convergence in each $L^1(Omega)$ for every bounded domain $Omega subset mathbb R^n,$ and hence weak $L^1$ convergence in $Omega.$ But $L^1(Omega)^* cong L^{infty}(Omega),$ which contains $C^{infty}_c(Omega).$
Going along these lines, we can check this directly. Suppose $f_n rightarrow f$ in $L^1_{mathrm{loc}}(mathbb R^n)$ and let $varphi in C^{infty}_c(mathbb R^n).$ Pick a bounded domain $Omega subset mathbb R^n$ such that $operatorname{supp} varphi subset Omega$ and note that $f_n rightarrow f$ in $L^1(Omega).$ Then we get,
$$ Lambda_{f_n}(varphi) - Lambda_f(varphi) = int_{mathbb R^n} (f_n - f)varphi ,mathrm{d}x = int_{Omega} (f_n - f)varphi ,mathrm{d}x leq lVert f_n - frVert_{L^1(Omega)} lVert varphi rVert_{L^{infty}_{Omega}} longrightarrow 0 $$
as $n rightarrow infty.$ Since $varphi$ was arbitrary, $Lambda_{f_n} rightarrow Lambda_f$ in $mathcal{D}'(mathbb R^n).$
answered Jan 4 at 12:36
ktoiktoi
2,3331616
add a comment |
Yes. The point is that convergence in $L^1_{mathrm{loc}}(mathbb R^n)$ implies convergence in each $L^1(Omega)$ for every bounded domain $Omega subset mathbb R^n,$ and hence weak $L^1$ convergence in $Omega.$ But $L^1(Omega)^* cong L^{infty}(Omega),$ which contains $C^{infty}_c(Omega).$
Going along these lines, we can check this directly. Suppose $f_n rightarrow f$ in $L^1_{mathrm{loc}}(mathbb R^n)$ and let $varphi in C^{infty}_c(mathbb R^n).$ Pick a bounded domain $Omega subset mathbb R^n$ such that $operatorname{supp} varphi subset Omega$ and note that $f_n rightarrow f$ in $L^1(Omega).$ Then we get,
$$ Lambda_{f_n}(varphi) - Lambda_f(varphi) = int_{mathbb R^n} (f_n - f)varphi ,mathrm{d}x = int_{Omega} (f_n - f)varphi ,mathrm{d}x leq lVert f_n - frVert_{L^1(Omega)} lVert varphi rVert_{L^{infty}_{Omega}} longrightarrow 0 $$
as $n rightarrow infty.$ Since $varphi$ was arbitrary, $Lambda_{f_n} rightarrow Lambda_f$ in $mathcal{D}'(mathbb R^n).$
answered Jan 4 at 12:36
ktoiktoi
2,3331616
Yes. The point is that convergence in $L^1_{mathrm{loc}}(mathbb R^n)$ implies convergence in each $L^1(Omega)$ for every bounded domain $Omega subset mathbb R^n,$ and hence weak $L^1$ convergence in $Omega.$ But $L^1(Omega)^* cong L^{infty}(Omega),$ which contains $C^{infty}_c(Omega).$
Going along these lines, we can check this directly. Suppose $f_n rightarrow f$ in $L^1_{mathrm{loc}}(mathbb R^n)$ and let $varphi in C^{infty}_c(mathbb R^n).$ Pick a bounded domain $Omega subset mathbb R^n$ such that $operatorname{supp} varphi subset Omega$ and note that $f_n rightarrow f$ in $L^1(Omega).$ Then we get,
$$ Lambda_{f_n}(varphi) - Lambda_f(varphi) = int_{mathbb R^n} (f_n - f)varphi ,mathrm{d}x = int_{Omega} (f_n - f)varphi ,mathrm{d}x leq lVert f_n - frVert_{L^1(Omega)} lVert varphi rVert_{L^{infty}_{Omega}} longrightarrow 0 $$
as $n rightarrow infty.$ Since $varphi$ was arbitrary, $Lambda_{f_n} rightarrow Lambda_f$ in $mathcal{D}'(mathbb R^n).$
answered Jan 4 at 12:36
ktoiktoi
2,3331616
answered Jan 4 at 12:36
ktoiktoi
2,3331616
answered Jan 4 at 12:36
ktoiktoi
2,3331616
answered Jan 4 at 12:36
ktoiktoi
2,3331616
2,3331616
add a comment |
add a comment |
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