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    Let f_{n} be the distribution \langle f_{n},\varphi\rangle=n(\varphi( \frac{1}{n})-\varphi(-\frac{1}{n})). I want to determine the \lim_{n\to\infty}\langle f_{n},\varphi\rangle.

    I thought about using L'Hospital's rule, since

    \displaystyle

\begin{aligned}

\lim_{n\to\infty} \langle f_{n},\varphi\rangle&=\lim_{n\to \infty}n\left(\varphi\left(\frac  {1}{n}\right)-\varphi\left(-\frac{1}{n}\right)\right)

\\

&=\lim_{n\to\infty} \frac{\left( \varphi\left(\frac{1}{n}\right)-\varphi\left(-\frac{1}{n}\right)\right)}{\frac  {1}{n}}

\end{aligned}

    And

    \displaystyle \lim_{n\to\infty}\frac{1}{n}= \lim_{n\to\infty}\left(\varphi \left(\frac{1}{n}\right)-\varphi\left(-\frac{1}{n}\right)\right)=0

    However, since for g(n)=1/n we have g'(n)=-1/n^2, we can see that the limit

    \displaystyle\lim_{n\to\infty} \frac{\left( \varphi'\left(-\frac{1}{n}\right)-\varphi'\left(\frac{1}{n}\right)  \right)}{\frac{1}{n^2}}

    will not be well defined. Does anyone have a better suggestion? Note that obviously \varphi\in\mathcal{D}.
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    Okay, I had another attempt. So, we have
    \begin{aligned} 	\lim_{n\to\infty} \langle f_{n},\varphi \rangle&= \lim_{n\to\infty}n \left(\varphi\left(\frac{1}{n} \right)-\varphi \left(-\frac{1}{n} \right)\right) 	\\ 	&=\lim_{n\to\infty}n( \langle \delta_{\frac{1}{n}},\varphi \rangle-\langle \delta_{-\frac{1}{n}},\varphi \rangle) 	\\ 	&=\lim_{n\to\infty}n \langle \delta_{\frac{1}{n}}-\delta_{-\frac{1}{n}},\varphi \rangle. 	\end{aligned}

    Recall that

     S_{N}[\delta](x)=\sum_{k=-N}^{N}e^{ikx}.

    Then

     \begin{aligned} 	S_{N}[\delta_{\frac{1}{n}}](x)&=\sum_{k=-N}^{N}e^{ik(x-\frac{1}{n})} 	\\ 	&=\sum_{k=-N}^{N}e^{ikx-\frac{ik}{n}} 	\\ 	&=\sum_{k=-N}^{N}e^{ikx}\sum_{k=-N}^{N}e^{-\frac{ik}{n}} 	\\ 	&=S_{N}[\delta](x)\left(\cos\left(\frac{k}{n} \right)-i\sin\left(\frac{k}{n} \right) \right), 	\end{aligned}

    and

    \begin{aligned} 	S_{N}[ \delta_{- \frac{1}{n}}](x)&=\sum_{k=-N}^{N}e^{ik(x+\frac{1}{n})} 	\\ 	&=\sum_{k=-N}^{N}e^{ikx+ \frac{ik}{n}} 	\\ 	&=S_{N}[ \delta](x) \left(\cos \left(\frac{k}{n} \right)+i\sin \left(\frac{k}{n} \right) \right). 	\end{aligned}

    Hence

     S_{N}[\delta_{\frac{1}{n}}](x)-S_{N}[ \delta_{- \frac{1}{n}}](x)=-2i\sin \left(\frac{k}{n} \right)\cdot S_{N}[ \delta](x).

    Thus

    \displaystyle \begin{aligned} \[ \lim_{n\to \infty}-2in \sin\left(\frac{k}{n} \right) \langle S_{N}[ \delta],\varphi \rangle 

\\

& = -2i \lim_{x \downarrow 0} \frac{\sin(xk)}{x} \langle S_{N}[ \delta],\varphi \rangle\qquad\text{where }x=\frac{1}{n} 

 \\ 

& =  -2 ik\langle S_{N}[ \delta],\varphi \rangle 	\\ 	&=2 \langle S_{N}[ \delta'], \varphi \rangle, 	\end{aligned}

    and so we can conclude that
     \lim_{n\to\infty} \langle f_{n},\varphi\rangle=2 \langle \delta',\varphi \rangle.
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    (Original post by RamocitoMorales)
    Let f_{n} be the distribution \langle f_{n},\varphi\rangle=n(\varphi( \frac{1}{n})-\varphi(-\frac{1}{n})). I want to determine the \lim_{n\to\infty}\langle f_{n},\varphi\rangle.

    I thought about using L'Hospital's rule, since

    \displaystyle

\begin{aligned}

\lim_{n\to\infty} \langle f_{n},\varphi\rangle&=\lim_{n\to \infty}n\left(\varphi\left(\frac  {1}{n}\right)-\varphi\left(-\frac{1}{n}\right)\right)

\\

&=\lim_{n\to\infty} \frac{\left( \varphi\left(\frac{1}{n}\right)-\varphi\left(-\frac{1}{n}\right)\right)}{\frac  {1}{n}}

\end{aligned}
    I think this is the more successful approach. Can't you write the above in terms of the derivative of the test function?
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    (Original post by RichE)
    I think this is the more successful approach. Can't you write the above in terms of the derivative of the test function?
    See the edit in the post above. I achieve a distributional result via the other approach.
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    (Original post by RamocitoMorales)
    See the edit in the post above. I achieve a distributional result via the other approach.
    Correct - but you could have got that much more quickly (in a line or two) from where I cut off the edit of your first approach.

    PS you may have a minus sign missing in your answer.
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    (Original post by RichE)
    Correct - but you could have got that much more quickly (in a line or two) from where I cut off the edit of your first approach.
    I agree that my method is far too long and complicated. Can you expand on what you mean by writing the derivatives of the test function?
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    (Original post by RamocitoMorales)
    I agree that my method is far too long and complicated. Can you expand on what you mean by writing the derivatives of the test function?

    \displaystyle

\lim_{n\to\infty} \frac{\left( \varphi\left(\frac{1}{n}\right)-\varphi\left(-\frac{1}{n}\right)\right)}{\frac  {1}{n}} =

 \lim_{n\to\infty} \frac{\left(\varphi\left(\frac{1  }{n}\right)-\varphi\left(0\right)\right)}{1/n} 

+

 \lim_{n\to\infty} \frac{\left( \varphi\left(0\right)-\varphi\left(-\frac{1}{n}\right)\right)}{1/n}
 
 
 
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