This claim would be equivalent to saying that all linear functionals on the topological vector space $C_c^{\infty}(\mathbb{R}^n)$ are continuous.
This is not true, indeed, on any infinite dimensional topological vector space $X$ over a field $K$ there is a linear functional which is discontinuous.
Indeed, let $X$ be an infinite dimensional topological vector space, and let $\left\{e_n\right\}$ be a vanishing $(e_n\to 0)$ infinite sequence of linearly independent vectors on $X$, which may be easily constructed using continuity of the scalar product map. Define a linear functional $f$ on $X$ as any linear extension of the map
$$f(e_n):=n $$
Then the sequence $\left\{e_n\right\}\subset X$ is such that $e_n\to 0$, but $f(e_n)\not\to 0$.
For instance, in $C_c^{\infty}(\mathbb{R})$ (but it is the same in higher dimensions), recall that we say that a sequence of test functions $\left\{\varphi_n\right\}$ converges to $0$ iff their supports $\operatorname{supp}\varphi_n$ are all contained on some compact set $K\subset \mathbb{R}$ and $\varphi_n^{(k)}\to 0$ uniformly on $K$ for each order of differentiation $k\in \mathbb{N}$.
EDIT: The above is the usual topology that is set in the space of test functions $C_c^{\infty}(\mathbb{R}^n)$. In order to discuss whether
a functional $f:C_c^{\infty}(\mathbb{R}^n)\to \mathbb{R}$ is continuous
we first need to fix a topology on $C_c^{\infty}(\mathbb{R}^n)$, so I
assume you are referring to this one.
We may start from a function $0\neq \varphi \in C_c^{\infty}(\mathbb{R})$ and then set
$$e_n(x):=\frac{1}{n}\varphi\left(x+\frac{1}{n}\right) $$
so that the $\left\{e_n\right\}$ are linearly independent (since their supports are disjoint), $\operatorname{supp}e_n\subset \operatorname{supp}\varphi+[-1,0)$ and $e_n^{(k)}=\frac{1}{n}\varphi^{(k)}\left(x+\frac{1}{n}\right)\to 0$ uniformly for all $k$ as $n\to +\infty$. Thus, $\left\{e_n\right\}$ is a vanishing sequence of linearly independent vectors, and the functional $f$ defined above in the general case is a discontinuous linear functional on $C_c^{\infty}(\mathbb{R})$.
On the other hand, this is basically a sequence of bump functions which quickly reduces in overall size as $n\to +\infty$ and slowly shifts towards the left. Can you imagine a reasonable linear functional which blows up on such a sequence of functions?
As $n\to +\infty$, nothing meaningful 'blows up' here. I think this goes to show that you can usually expect your distributions to be continuous.
First of all, let me review some generalities: A "distribution" $T\in\mathcal{D}^{\prime}(\mathbb{R}^{4})$ on $\mathbb{R}^{4}$ is defined to be a linear and continuous functional of the form
$$T:\mathcal{D}(\mathbb{R}^{4})\to\mathbb{C}$$
where $\mathcal{D}(\mathbb{R}^{4})$ is some space of test functions. Usual choices are $C_{c}^{\infty}(\mathbb{R}^{4})$, $C^{\infty}(\mathbb{R}^{4})$ and $\mathcal{S}(\mathbb{R}^{4})$ equipped with some specific topologies, where in the latter case these distributions are usually called "tempered distributions".
Now, for every (nice enough, i.e. locally intergrable) function $f$, you can define a distribution via
$$T_{f}(\varphi):=\int_{\mathbb{R}^{4}}\varphi(x)f(x).$$
However, not every distribution is of this form. A distribution $T$ for which there is a $f$ such that $T=T_{f}$ is called "regular".
Now to your questions:
- If the quantum field $\phi$ would be regular, then the correct way of writing it would be
$$\Phi(\varphi)=\int_{\mathbb{R}^{4}}\varphi(x)\phi(x),$$
where $x=(t,\vec{x})$ denotes a 4-vector, i.e. an element of $\mathbb{R}^{4}$. So, you don't have to take two test functions one for the space and one for the time component. (not every function $f\in\mathcal{D}(\mathbb{R}^{4})$ can be factorized as $f(x)=f_{1}(t)f_{2}(\vec{x})$.)
- However, the operator-valued distribution $\phi$, i.e. the quantum field, is in general$^{1}$ not regular, i.e. it can not be written as a function. What Folland means with "maintaining the notational fiction that distributions are functions" is that you write all the equations as if it the distribution $\Phi$ was regular, in order to simplify notation. The situation is actually very similar to the delta function: The delta distribution is defined by
$$\delta(\varphi)=\varphi(0)$$
Now, one can show that this distribution is not regular, i.e. there does not exist a function $\delta$ (denoted by the same symbol for simplicity) such that
$$\delta(\varphi)=\int_{\mathbb{R}^{4}}\delta(x)\varphi(x)\,\mathrm{d}x=\varphi(0)$$
However, still, it is often very useful to write down equation in terms of the fictional function $\delta$, in order to keep the notation simple. These equations make then only sense if we integrate over them. In other words, one views the expression
$$\int_{\mathbb{R}^{4}}\delta(x)\varphi(x)\,\mathrm{d}x$$
just as a notation for the actual distribution $\delta$. The same is meant in the context of quantum field theory. The quantum field is a operator-valued distribution $\Phi$, which is not regular, i.e. there does not exist a function $\phi$. However, to simplify notation, we write down all equation in terms of this "fictional" function $\phi$. These equation then have to be interpreted in the sense that we have to integrate against a test function to recover the actual well-defined object.
As to your confusion with the creation and anihilation operator: Yes, $a,a^{\dagger}$ are also distributions. What you really do is to define a quantum field as
$$\Phi(f)\propto a(f)+a^{\dagger}(f)$$
for all test functions $f$ (This formalism is sometimes called "Segal quantization"). Then, you introduce the fictional field $\phi$, i.e. you formally write
$$\Phi(f)=\int\,f(x)\varphi(x)$$
where, as explained above, $\varphi$ does not exist as an actual function. This, in turn, leads to the formula you used, i.e. using ficitional functions $a$ and $a^{\dagger}$ for the distributions denoted with the same symbol.
$^{1}$ More formally, one can show, that there does not exist an operator-valued function satisfying the Wightman axioms of quantum field theory. From the physical point of view, this could have been expected and is basically a consequence of the uncertainty principle: If you measure your field at some point, then this would cause very large fluctuations of momentum, which in turn cannot lead to a well-defined operator. This is also the reason why one more formally considers distributions functions in order to defined quantum fields.
Best Answer
In which book by Folland did you read this?
I agree with you that continuity of $T'$ seems not be needed for continuity of $T$ extended to $\mathcal{D}'.$ However, I think that it is needed for $TF$ to be a distribution, i.e. for $TF$ to be continuous:
Let $\varphi_j \to \varphi$ in $C_c^\infty$ and assume that $T'$ is continuous. Then, $ \langle TF, \varphi_j \rangle = \langle F, T'\varphi_j \rangle \to \langle F, T'\varphi \rangle = \langle TF, \varphi \rangle ,$ i.e. $TF$ is continuous.