TITLE: Why delta function is a tempered distribution?
QUESTION [7 upvotes]: Tempered distribution $T\in \mathcal{S}'$ is a continuous linear functional defined on the space $\mathcal{S}$ of infinitely differentiable test functions $f(x)$ on $\mathbf{R}$ with finite norm
$$
||f||_{r,s} = \sum_{k,|k|<r}\sum_{l,|l|<s} \mathop{\mathrm{sup}}_x \left|x^k f^{(l)}\right| < \infty,
$$
for all $r,s$ ($f^{(l)}$ is the $l$-th derivative).
A theorem states, that all tempered distributions have the form
$$
T(f) = \sum_{0\leq|k|\leq s}\int F_k(x) f^{(k)}(x) dx,
$$
with some continuous functions $F_k$ bound as
$$
|F_k(x)| \leq C_k(1+|x|^j),
$$
with some $C_k$ and $j$ depending on $k$.
According to the theorem the delta function distribution defined as
$$
T_\delta(f) = \int \delta(x) f(x) dx = f(0),
$$
is not a tempered distribution ($\delta(x)$ is not a bound continuous function without doubt. I write it here in the integral just for the sake of traditional notations). Question, why it is not a tempered distribution? Naively, it is linear, defined an all functions in $\mathcal{S}$, and seems to be continuous?
Note, that there were couple of questions here that asked, why $T_\delta$ is not induced by some good function $F_k$ — this is relatively obvious.  The question is, why $T_\delta$ is not a good distribution in $\mathcal{S}'$. For example, it is a good distribution if a smaller set of test functions is used, that is the set $\mathcal{D}$ of functions with finite support.

Definitions of $S$ and $S'$ can be found, e.g. in PCT, Spin and Statistics, and All That by R.F. Streater, A.S. Wightman
Theorem is mentioned in the same book, and seems to be proven in L.Schwartz Théorie des distributions; or L.Gårding and J.Lions "Functional Analysis" Nuovo Cimento Suppl., 14, 9 (1959).

(Question title edited after the proper answer arrived)

REPLY [3 votes]: Dirac delta is a tempered distribution. Indeed, for each $f \in \mathcal{S}(\mathbb{R})$,
\begin{align*}
\delta(f)
= f(0)
= -\int_{\mathbb{R}} \mathbf{1}_{(0,\infty)}(x) f'(x) \,dx
= \int_{\mathbb{R}} \max\{0,x\} f''(x) \,dx.
\end{align*}
Here $\mathbf{1}_{(0,\infty)}(x)$ is the indicator function of $(0, \infty)$ and we applied integration by part at the last step. So, the Dirac delta satisfies the theorem with the continuous function $F_2(x) = \max\{0,x\}$ satisfying the bound $|F_2(x)| \leq |x|$.
As a remark, this computation simply tells that $\frac{d^2}{dx^2} \max\{0, x\} = \delta(x)$ in distribution sense. This is what @Blazej is mentioning in his/her comment.