How to prove that the Dirac delta is not a function? I am currently taking a course on test functions and distributions and my task is to prove that the Dirac delta is not a function. 
Furthermore, I would also like to prove that it is continuous as a distribution in the sense that the inequality :
$|\delta_{a}(\phi)| \leq c\|{\phi}\|_{C^k}$, for $c>0$ and $\phi \in C_{0}^{\infty}$ holds.
Thank you!
 A: Suppose that we have a function $\psi(x)$ so that
$$
\delta(\phi)=\int_{\mathbb{R}}\psi(x)\phi(x)\,\mathrm{d}x\tag{1}
$$
Define
$$
\begin{align}
\eta(x)&=\left\{\begin{array}{}
0&\text{if }|x|\le\frac12\\
\frac{\displaystyle e^{\frac1{1-2|x|}}}{\displaystyle e^{\frac1{1-2|x|}}+e^{\frac1{|x|-1}}} &\text{if }\frac12\lt|x|\lt1\\
1&\text{if }|x|\ge1
\end{array}\right.\tag{2a}\\[6pt]
\eta_k(x)&=\eta\!\left(2^{-k} x\right)\tag{2b}
\end{align}
$$
Then $\eta_k\in C^\infty$, $\eta_k(x)=0$ for $|x|\le2^{k-1}$, and $\eta_k(x)=1$ for $|x|\ge2^k$. Since $C^\infty$ is dense in $L_\text{loc}^1$, for any $\epsilon>0$, we can choose a $\psi_\epsilon\in C^\infty$ so that
$$
\begin{align}
\|(\psi-\psi_\epsilon)(1-\eta)\|_{L^1}&\le\epsilon/2\tag{3a}\\
\|(\psi-\psi_\epsilon)(\eta_k-\eta_{k+1})\|_{L^1}&\le2^{-k-2}\epsilon\quad\text{for }k\ge0\tag{3b}
\end{align}
$$
which means
$$
\|\psi-\psi_\epsilon\|_{L^1}\le\epsilon\tag{3c}
$$
Furthermore, let $\sigma$ be the signum function, then since $\psi_\epsilon\in L_\text{loc}^\infty$, we can choose a $\sigma_\epsilon\in C^\infty$ so that $|\sigma_\epsilon|\le1$ and
$$
\begin{align}
\|(\sigma\circ\psi-\sigma_\epsilon)(1-\eta)\psi_\epsilon\|_{L^1}&\le\epsilon/2\tag{4a}\\
\|(\sigma\circ\psi-\sigma_\epsilon)(\eta_k-\eta_{k+1})\psi_\epsilon\|_{L^1}&\le2^{-k-2}\epsilon\quad\text{for }k\ge0\tag{4b}
\end{align}
$$
which means
$$
\|(\sigma\circ\psi-\sigma_\epsilon)\psi_\epsilon\|_{L^1}\le\epsilon\tag{4c}
$$
Since $\|\sigma\circ\psi-\sigma_\epsilon\|_{L^\infty}\le2$, we get
$$
\begin{align}
\|(\sigma\circ\psi-\sigma_\epsilon)\psi\|_{L^1}
&\le\|(\sigma\circ\psi-\sigma_\epsilon)\psi_\epsilon\|_{L^1}+\|(\sigma\circ\psi-\sigma_\epsilon)(\psi-\psi_\epsilon)\|_{L^1}\tag{5a}\\
&\le\|(\sigma\circ\psi-\sigma_\epsilon)\psi_\epsilon\|_{L^1}+\|\sigma\circ\psi-\sigma_\epsilon\|_{L^\infty}\|\psi-\psi_\epsilon\|_{L^1}\tag{5b}\\
&\le3\epsilon\tag{5c}
\end{align}
$$
Define $\phi_k(x)=\sigma_\epsilon(x)\left(\eta(2^{-k} x)-\eta(2^{-k-1} x)\right)$. Then we have
$$
\begin{align}
0
&=\delta\left(\sum_{k=m}^n\phi_k\right)\tag{6a}\\
&=\sum_{k=m}^n\int_{\mathbb{R}}\psi(x)\,\phi_k(x)\,\mathrm{d}x\tag{6b}\\
&=\sum_{k=m}^n\int_{\mathbb{R}}\psi(x)\,\sigma_\epsilon(x)\left(\eta(2^{-k} x)-\eta(2^{-k-1} x)\right)\mathrm{d}x\tag{6c}\\
&=\int_{\mathbb{R}}\psi(x)\,\sigma_\epsilon(x)\left(\eta(2^{-m} x)-\eta(2^{-n-1} x)\right)\mathrm{d}x\tag{6d}\\
&\ge\int_{\mathbb{R}}|\psi(x)|\left(\eta(2^{-m} x)-\eta(2^{-n-1} x)\right)\,\mathrm{d}x\\
&-\int_{\mathbb{R}}|(\sigma\circ\psi(x)-\sigma_\epsilon(x))\psi(x)|\left(\eta(2^{-m} x)-\eta(2^{-n-1} x)\right)\mathrm{d}x\tag{6e}\\
&\ge\int_{2^m\lt|x|\lt2^n}|\psi(x)|\,\mathrm{d}x-3\epsilon\tag{6f}
\end{align}
$$
Explanation:
$\text{(6a)}$: $\phi_k\in C_c^\infty$ and is $0$ in a neighborhood of $0$
$\text{(6b)}$: apply $(1)$
$\text{(6c)}$: apply the definition of $\phi_k$
$\text{(6d)}$: compute the sum of the $\eta_k$
$\text{(6e)}$: $\psi\sigma_\epsilon=|\psi|-(\sigma\circ\psi-\sigma_\epsilon)\psi$
$\text{(6f)}$: apply $(5)$
Since $\epsilon$ was arbitrary, $(6)$ implies that on any annulus centered at $0$, $\psi$ is $0$. That is, $\psi(x)=0$ for all $x\ne0$. Since the value of a function at a single point does not affect the integral of that function, $(1)$ would imply that
$$
\delta(\phi)=0\tag7
$$
for all $\phi$. Since $(7)$ is false, $(1)$ must be false.
A: Although this question is very old and it already has an accepted answer that seems to be correct, I will add another one, much shorter and slightly easier. I will do so also because this question has many views and this basic fact seems to be easily overlooked in many standard texts providing an introduction to distributions (I guess that the two things are actually intertwined).
What follows is taken from Example 1.1 in G. Grubb, Distributions and operators, Graduated texts in mathematics, Springer Verlag (2009). This is a variant of the approach suggested in the comments below the OP.
Recall that $\delta$ is the distributional derivative of the Heaviside function $$H(x) := \begin{cases} 1 & x > 0,\\ 0 & x \leq 0.\end{cases}$$
In other words, $\delta(\phi) = -\int_{\mathbb R} H \phi' \,dx = \phi(0)$.
The continuity of $\delta$ follows because $|\delta(\phi)| = |\phi(0)| \leq \|\phi\|_{C^0(\mathbb R)}$ (see section 3.1 in Grubb, or Theorem 6.8 in Rudin's Functional Analysis).
Suppose there exists $v \in L^1_{\rm loc}(\mathbb R)$ such that, for all $\phi \in C^\infty_c(\mathbb R)$, it holds $$\delta(\phi) = \int_{\mathbb R} v(x) \phi(x) \,dx. \quad \tag{$*$}$$ Let $\phi \in C^\infty_c(\mathbb R)$ be such that $\phi(0) = 1$ and, for $N \in \mathbb N$, set $\phi_N(x) := \phi(Nx)$. Note that, for all $N$, $\max |\phi(x)| = \max|\phi_N(x)|$, and that when $\phi$ is supported in $[-R,R]$, $\phi_N$ is supported in $[-R/N, R/N]$. By the theorem of Lebesgue,
$$\int_{\mathbb R} v \phi_N \,dx \to 0 \quad \mbox{as } N \to \infty.$$
On the other hand,
$$-\int_{\mathbb R} H\phi_N' \,dx = -\int_0^{+\infty} N \phi'(Nx) \,dx = -\int_0^{+\infty} \phi'(y) \,dy = \phi(0) = 1,$$
a contradiction. Hence, ($*$) does not hold for the sequence $\{\phi_N\} \subset C^\infty_c(\mathbb R)$, and in turn $v$ cannot exist.
