In the space of test functions, does $f_j\to f$ imply $f_j-f\to 0$?

Question

Let $\mathcal D(\Omega)$ be a set of test functions on open $\Omega\subset \mathbb R$, and let $\{f_j\}_{j=1}^\infty\subset\mathcal D(\Omega), f\in\mathcal D(\Omega).$

The definition of $f_j\to f$ in $\mathcal D(\Omega)$ is :

(i) There is a compact set $K\subset\Omega$ s.t. $\forall j, \mathrm{supp} f_j\subset K$

(ii) For all multiindex $\alpha$, $\displaystyle\lim_{j\to\infty}\ \sup_{x\in K}\|D^\alpha f_j(x)-D^\alpha f(x)\|=0.$

I want to know whether or not $f_j\to f$ in $\mathcal D(\Omega)$ implies $f_j-f\to 0$ in $\mathcal D(\Omega).$

So, suppose $f_j\to f$ in $\mathcal D(\Omega)$. To see $f_j-f\to 0$ in $\mathcal D(\Omega)$, we have to find a compact $K'\subset\Omega$ s.t.

(i)' $\forall j, \mathrm{supp} (f_j-f)\subset K'$

(ii)' For all multiindex $\alpha$, $\displaystyle\lim_{j\to\infty}\ \sup_{x\in K'}\|D^\alpha (f_j-f)(x)-D^\alpha\ 0\|=\lim_{j\to\infty}\ \sup_{x\in K'}\|D^\alpha f_j(x)-D^\alpha f(x)\|=0.$

I'm wondering how I find $K'.$

Now, $f\in\mathcal D(\Omega)$ gives a compact $K_0\subset\Omega$ s.t. supp$f=K_0.$

Let $K':=K\cup K_0.$ Then, (i)' holds since supp$(f_j-f)\subset$ supp$f_j\cup$ supp $f\subset K'$.

But I cannot see (ii)''.

Can we see $\lim_{j\to\infty}\ \sup_{x\in K\cup K_0}\|D^\alpha f_j(x)-D^\alpha f(x)\| =0$ ?

Since $K\subset K\cup K_0$, we can see $\sup_{x\in K\cup K_0}\|D^\alpha f_j(x)-D^\alpha f(x)\|\geqq \sup_{x\in K}\|D^\alpha f_j(x)-D^\alpha f(x)\|$ but I need opposite direction inequality...

Answer 1

As already mentioned, in the definition of the convergence $f_j\rightarrow f$ in $\mathcal{D}(\Omega)$ we somehow have to include that $\text{supp}f\subset K$, where $K$ is as in (i). There are two ways to fix this this.

1. Include $\text{supp}f\subset K$ in (i) of your definition.
2. Require in (ii) that $D^\alpha f_j$ converges to $D^\alpha f$ uniformly on $\Omega$ as $j\rightarrow\infty$ for all $\alpha\in\mathbb{N}$, i.e. $\lim\limits_{j\rightarrow\infty} \sup_{x\in\Omega} \Vert D^\alpha f_j(x) - D^\alpha f(x) \Vert = 0$. (Although it is enough to require uniform convergence on $K$ and point wise convergence on $\Omega\setminus K$, since then it follows that $f(x) = 0$ for all $x\in\Omega\setminus K$ and thus $\text{supp}f \subset K$).

Otherwise it wouldn't really make sense if the support of $f$ is bigger than the one from the $f_j$'s. And with this definition your equivalence of the convergences should be clear.