I'm a starter in Fourier analysis...

in the following examples for $\phi\in C_0^{\infty}(\mathbb{R}) $ with $0 \notin \text{supp} \ \phi$ and $\phi \neq 0$ i like to show that

$$(1) j^{-1}\phi(x-j), \ \ \ (2) j^{-p}\phi(jx) \ \ \ (3) e^{-j}\phi(jx)$$

($p$ some number), all have the property that they converge to $0$ in $C_0^{\infty}(\mathbb{R})$. So that is verifying that for all $k$ that $(\phi_j^{(k)}(x))\to 0$ for all $k$.(in case $(1)$ even uniformly apparently.)

To me this seems Obvious in all cases since if supp $\phi$ is compact, it is bounded...and $\text{supp} \ \phi_j^{(k)} \subset \text{supp} \ \phi_j$ so that we eventually move out of its support in all cases $(1),(2),(3)$

So i dont fully see what has to be verified here...

A sequence of test-functions $(\phi_j)\subset C_0^{\infty}(\mathbb{R}) $ converges to $\phi_0$ in the following sense that for an increasing sequence of compact sets $(K_l)$ with $\bigcup K_l = \mathbb{R}$ there exists $l$ s.t. $ \text{supp} \ \phi_j \subset K_l$ for all j and we have

$\sup_{x\in \mathbb{R}} | \phi_0^{(k)}(x)- \phi_j^{(k)}(x)|\to 0$

for all $k\in \mathbb{N}$

  • $\begingroup$ DO i need to be extra careful in one of these examples..? For (2) deravitive is $j^{k-p}\phi^{(k)}(jx)$...but that also converges quite clearly to 0. $\endgroup$
    – DinkyDoe
    Commented Feb 13, 2013 at 14:03
  • $\begingroup$ yes, but then since $\phi^{(k)}(jx) = 0$ since $jx \notin \text{supp} \ \phi_j^{(k)} \subset \text{supp} \ \phi$ for sufficiently large j. $\endgroup$
    – DinkyDoe
    Commented Feb 13, 2013 at 14:14
  • $\begingroup$ I edited my question...so the topology u were probably already referring to in your last comment. $\endgroup$
    – DinkyDoe
    Commented Feb 13, 2013 at 14:22
  • $\begingroup$ Sorry..im afraid that is actually the topology i meant. Ill even re-edit my question. Thanks for pointing out. $\endgroup$
    – DinkyDoe
    Commented Feb 13, 2013 at 14:41
  • $\begingroup$ Okay, I've cleared out my comments. An answer in a bit. $\endgroup$ Commented Feb 13, 2013 at 14:46

1 Answer 1


Using the stated topology on test functions, the only one of the three cases where the sequence of functions converge to 0 is case (3).

Case (1) fails because if $x_0$ is such that $\phi(x_0) \neq 0$, we must have that $\phi((x_0 + j) - j) \neq 0$. So for the sequence of functions $\phi_j(x) = j^{-1} \phi(x - j)$, there does not exist any compact set $K$ such that $\sup \phi_j \subseteq K$ for all $j$. ($K$ will need to contain the sequence $\{x_0 + j\}$ which has no converging subsequence).

Case (2) fails because the lack of uniform convergence. Let $R = \sup_{x \in \mathrm{supp} \phi} |x|$. Then $[-R,R]$ is a compact set that contains all the supports of $\phi_j(x) = j^{-p} \phi(jx)$. But if $k > p$ we have that (setting $\phi_0 \equiv 0$)

$$ \sup_{x\in\mathbb{R}} | \phi_0^{(k)} - \phi_j^{(k)}| = \sup_{x \in \mathbb{R}} | j^{k-p} \phi^{(k)}(jx)| \geq j^{k-p} \sup_{x\in\mathbb{R}} | \phi^{(k)}(x)| $$

the right hand side increases without bound as $j$ increases, and so the sequence $\phi_j$ cannot converge to 0.

For case(3), the compact set $[-R,R]$ as before will still contain the supports of all $\phi_j(x) = e^{-j}\phi(jx)$. And now we have that

$$ \sup_{x\in\mathbb{R}} | \phi_0^{(k)} - \phi_j^{(k)}| = \sup_{x \in \mathbb{R}} | e^{-j} j^k \phi^{(k)}(jx)| = e^{-j} j^{k} \sup_{x\in\mathbb{R}} | \phi^{(k)}(x)| $$

For each fixed $k$, we have that $\lim_{j\to\infty} e^{-j} j^k = 0$. This shows that the condition for convergence is in fact verified.

  • $\begingroup$ Thank you kindly for your time. $\endgroup$
    – DinkyDoe
    Commented Feb 13, 2013 at 14:58

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