Translation operator and test function on $\mathbb{R}$

analysisdistribution-theoryfunctional-analysisreal-analysis

Consider the translation operator $\tau_h, h\in \mathbb{R}$, on the space of test functions $\mathcal{D}(\mathbb{R})$ on $\mathbb{R}$. I would like to prove that for any $\phi \in \mathcal{D}(\mathbb{R})$, we have
$$ \frac{1}{h} \cdot (\phi – \tau_h \phi) \, \to \phi' \quad \text{in $\mathcal{D}(\mathbb{R})$} \quad \text{as $h\to 0$.}$$

I would like to use the characterization of the convergence in $\mathcal{D}(\mathbb{R})$ via compact subsets of $\mathbb{R}$ I'm stucked in how to approach the problem.

Any suggestions? Thanks in advance!

Best Answer

I’m not sure about your conventions, so let’s say $(\tau_h\phi)(x)=\phi(x-h)$, so that the result is correct.

Let $F_h=\frac{\phi-\tau_h\phi}{h}-\phi’$ for $h \neq 0$.

We want to show that in $\mathcal{D}(\mathbb{R})$, $F_h \rightarrow 0$ as $h \rightarrow 0$. This means showing two things:

  1. that there exists a compact subset $K \subset \mathbb{R}$ and $h_0>0$ such that for each $h$ such that $0<|h| < h_0$, $F_h$ has support in $K$.

  2. that for every $n \geq 0$, $\sup_{x \in \mathbb{R}}\,|F_h^{(n)}(x)| \rightarrow 0$ as $h \rightarrow 0$.

For 1), just take $K=L+[-1,1]$ where $L$ is the support of $\phi$ (since $\phi \in \mathcal{D}(\mathbb{R})$) and $h_0=1$.

Let’s see 2). For each $n \geq 0$, for each real number $x$, for each $h \neq 0$, $F_h^{(n)}(x)=\frac{\phi^{(n)}(x)-\phi^{(n)}(x-h)-h\phi^{(n+1)}(x)}{h}=-h^{-1}(\phi^{(n)}(x-h)-\phi^{(n)}(x)-(-h)\phi^{(n+1)}(x))$, so by Taylor, $$F_h^{(n)}(x)=-h^{-1}\int_x^{x-h}{(x-h-t)\phi^{(n+2)}(t)dt}=-h^{-1}\int_0^{-h}{(-t-h)\phi^{(n+2)}(x+t)dt},$$ thus $$F_h^{(n)}(t)=h^{-1}\int_h^0{t\phi^{(n+2)}(x-h+t)dt}=-h\int_0^1{t\phi^{(n+2)}(x-h+ht)dt}.$$

Finally, $|F_h^{(n)}(x)| \leq |h| \|\phi^{(n+2)}\|_{\infty}$, hence the conclusion.