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Sequence of continuous functions $(f_k)$ which converge to $f$ in $C(\Bbb R)$. Does it imply that $f_k$ will converge to $f$ in $\mathcal D'(\Bbb R)$ as well? Or we need something more to make this assertion?

[Here $\mathcal D'(\Bbb R)$ is a linear continuous functional on $\mathcal D(\Bbb R)$ ; collection of all linear maps: from $\mathcal D(\Bbb R)$ to $\Bbb R$)

and we say $f\in$ $\mathcal D'(\Bbb R)$, is continuous in the sense that for all $\phi_n\to \phi$ in $\mathcal D(\Bbb R)$ => $(f,\phi_n)\to(f,\phi)$ in $\mathcal D'(\Bbb R)$]

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what do you mean exactly by convergence in $C(R)$ ? uniform convergence on compacts ? –  Glougloubarbaki May 20 '13 at 21:38
    
What type of convergence are you taking in $C(\mathbb{R})$? –  Tomás May 20 '13 at 21:38
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But you have to especify what you mean by convergence in this space. Also, in some few lines after it, you use the space $\mathcal{D}(\Omega)$. Its does not make sense, the function $f=1$ for all $x$ is in $C(\mathbb{R})$ but does not belong to $\mathcal{D}(\mathbb{R})$ –  Tomás May 20 '13 at 21:44
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I - There is not motive a priori for $f\in C(\mathbb{R})$ implies that $f\in \mathcal{D}'(\mathbb{R})$. II - To say that $f_k\to f$ in $C(\mathbb{R})$ you have to especify what this mean (or in other words you have to define it). –  Tomás May 20 '13 at 22:04
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@SaraTancredi every continuous function is a distribution, but you still need to specify what you mean by $f_k \rightarrow f$ in $C(\mathbb{R})$ –  Glougloubarbaki May 20 '13 at 22:32
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1 Answer

It seems that the type of convergence considered is: $f_n\to f$ in $C(\mathbb{R})$ if for every fixed compact $K\subset \mathbb{R}$ we have that $f_n\to f$ uniformly in $K$. By using this definition we have that $f_n\to f$ as a distribution, i.e. if we define $T_n(\phi)=\int_\mathbb{R} f_n\phi$ and $T(\phi)=\int_\mathbb{R}f\phi$, $\phi\in \mathcal{D}(\mathbb{R})$ then $T_n\to T$ in $\mathcal{D}'(\mathbb{R})$

To prove the claim in the last paragraph, let $K=\overline{\operatorname{support}(\phi)}$, where by definition $K$ is a compact set. Hence, $f_n\to f$ uniformly in $K$, which implies that

$T_n(\phi)=\int_\mathbb{R}f_n\phi=\int_K f_n\phi\to \int_Kf\phi=\int_\mathbb{R} f\phi=T(\phi)$ for every $\phi\in \mathcal{D}(\mathbb{R})$

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