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)
Best Answer
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.