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

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...

• I'd say that due to the convergence we necessarily have $\text{supp}f \subset K$. Jun 2, 2023 at 11:07
• Do you mean that supp$f\subset K$ should be in the supposition ? @stange
– daㅤ
Jun 2, 2023 at 19:17

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.