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Find the distributional limit of $f_n = e^{x/n}$.

That means that I have to find the limit of $T_{f_n}$. I proceeded as follows. Let $\phi \in \mathcal D$ be a test function. Then: $$\langle T_{f_n}, \phi \rangle = \int_{\mathbb R} e^{\frac xn}\phi(x)\,\mathrm dx = -n\int_{\mathbb R} e^{x/n} \phi'(x)\,\mathrm dx$$ where I integrated by parts and exploited the fact that $\phi$ has compact support. From the same fact it follows that the integral is finite (since we also have that $e^{x/n}$ is continuous), and therefore $$\langle T_{f_n}, \phi\rangle \to +\infty$$

I don't know how to interpret this result. In the other exercises, the result was $0$, which is the zero distribution. Here, do I have to conclude that the limit does not exist?

If the limit was a non-zero finite number, should I also say that the limit does not exist? Since a non-zero constant wouldn't identify a distribution.

EDIT: In the sentence above lay my misunderstanding. I was confusing distributions and test functions. A constant $f \equiv c \in \mathbb R$ does indeed identify a distribution, namely $$\langle T_f, \phi \rangle = \int_{\mathbb R} c \cdot \phi(x)\,\mathrm dx \in \mathbb R$$ since $f$ is locally integrable on $\mathbb R$.

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    $\begingroup$ Note that the integral depends on $n$ and can go to $0$. $\endgroup$ – N. S. Jul 7 '16 at 12:36
  • $\begingroup$ You can use dominated convergence. $\endgroup$ – user223391 Jul 8 '16 at 22:39
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Hint $\phi$ has compact support and $e^{\frac{x}{n}} \to 1$ uniformly on compact sets.

P.S. Note that your integration by parts approach cannot help: Indeed, since $e^{x/n} \to 1$ on $supp(\phi')$, we have $$\int_{\mathbb R} e^{x/n} \phi'(x) dx \to \int_{\mathbb R} \phi'(x) dx =0$$ with the last equality following from the fact that the antiderivative of $\phi'$ has compact support.

This means that the limit you get is of the type $\infty \cdot 0$.

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  • $\begingroup$ I'm sorry I don't understand. Even if $e^{x/n} \to 1$ the integral is zero only if $\phi \equiv 0$. $\endgroup$ – rubik Jul 8 '16 at 20:02
  • $\begingroup$ @rubik If $e^{x/n} \to 1$ uniformly on the support of $\phi$ then $\int_{\mathbb R} e^{x/n} \phi(x)$ converges to.... $\endgroup$ – N. S. Jul 8 '16 at 22:22
  • $\begingroup$ @rubik Check the P.S. for clarification. $\endgroup$ – N. S. Jul 8 '16 at 22:36
  • $\begingroup$ I think I understand. Since $e^{x/n} \to 1$ uniformly, by exchanging limit and integral I get to $$\langle T_{f_n}, \phi \rangle \to \int_{\mathbb R}1 \cdot \phi(x)\,\mathrm dx = \langle T_f, \phi \rangle$$ where $f \equiv 1$. Is it correct? $\endgroup$ – rubik Jul 11 '16 at 17:02
  • $\begingroup$ @rubik Yes. That is correct. $\endgroup$ – N. S. Jul 11 '16 at 17:11

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