Proof that $\delta_0 \not\in C^0(\mathbb{ R}^n)$ I'm doing a course in PDEs, and this has come up as a problem. I've attempted a proof, but I've run into the issue that according to the professor, the only property that may be used is the fact that $\langle \delta_0, \phi \rangle = \int_{\mathbb{R}^n} \delta_0(x) \phi(x) dx = \phi(0)$. 
The only hint he was willing to give was to assume that $\delta_0$ is $C^0(\mathbb{R}^N)$, and show a contradiction. The problem that I'm having is that from what I've been able to find, every property of continuous functions requires the concept of a norm, something which $d_0$ doesn't have.
From the tone of his answer, he made it sound as though using any concept of a norm is therefore unacceptable, even in the context of a proof by contradiction, but I don't see how it could even be possible to make a statement about continuity without having some notion of a norm.
This was my attempt at a proof (the definition of \delta_0 that I use here was given in class):
Assume by absurd that $\delta_0 \in C^0(\mathbb{R}^N)$. Then given that $
\delta_0 \equiv \lim_{\epsilon \to 0} f_\epsilon$, where 
\begin{equation}
    f_\epsilon(x) =\begin{cases}
0 \quad & x \not\in B_\epsilon(0)\\
\frac{1}{\epsilon} \quad & x \in B_\epsilon(0)
\end{cases}
\end{equation}
and $B_\epsilon(0)$ is a ball with radius $\epsilon$ centered at $x=\vec{0}$, and given
$\tilde{\epsilon}>0$, there exists $\tilde{\delta}>0$ such that $||x-y||<\tilde{\delta}$ in the Euclidean norm for $x,y\in\mathbb{R}^N$ implies that $|\lim_{\epsilon \to 0}f_{\epsilon}(x) - \lim_{\epsilon \to 0}f_{\epsilon}(y)| < \tilde{\epsilon}$. 
Take $y = \vec{0}$, $x \in \mathbb{R}^N$\ $\vec{0}$. In such a case, $$||x-y||=||x-\vec{0}||=||x||<\tilde{\delta}$$ $$\implies |\lim_{\epsilon \to 0}f_{\epsilon}(x) - \lim_{\epsilon \to 0}f_{\epsilon}(\vec{0})|=|0 - \lim_{\epsilon \to 0} \frac{1}{\epsilon}| = \lim_{\epsilon \to 0} \frac{1}{\epsilon} <\tilde{\epsilon}$$
But,  $\lim_{\epsilon \to 0} \frac{1}{\epsilon}$ is clearly unbounded. Thus, the assumption that $\delta_0 \in C^0(\mathbb{R}^N)$ leads to a contradiction, therefore $\delta_0 \not\in C^0(\mathbb{R}^N) $.
 A: $\delta_0 \in C^0(\mathbb{R}^n$) implies $\delta_0 \in L^2(B_1(0))$, where $B_1(0)$ denotes the ball of radius $1$ around the origin. This is true since any continuous function is bounded on a compact set and therefore 
$$
\int_{B_1(0)} |\delta_0(x)|^2 \; dx \leq \sup_{x\in \overline{B_1(0)}} |\delta_0(x)|^2 \mu(B_1(0)) \leq C^2  \mu(B_1(0)) < \infty
$$
Asssume $\delta_0 \in L^2(B_1(0))$. Then, 
$$
F: L^2(B_1(0)) \mapsto \mathbb{R},f \mapsto \int_{B_1(0)} \delta_0(x)f(x)
$$
defines a continuous, linear functional on $L^2(B_1(0))$ by the Cauchy-Schwarz inequality ( $|F(f)|\leq ||\delta_0||_{L^2(B_1(0))}||f||_{L^2(B_1(0))}$). Consider the sequence 
$$
f_k(x)=\sqrt k e^{-k|x|^2}\mathbb{1}_{B_1(0))}(x)
$$
Then $f_k \to 0$ in $L^2(B_1(0))$ and hence our continous, linear functional should go to $0$, i.e. $F(f_k) \to 0$. But 
$$
F(f_k)=\int_{B_1(0)} \delta_0(x)f_k(x)=\int_{\mathbb{R}^n} \delta_0(x)f_k(x)=\sqrt{k} \to \infty 
$$ 
Hence $\delta_0 \notin L^2(B_1(0))$ and therefore $\delta_0 \notin C^0(\mathbb{R}^n)$.
A: You may rephrase it in the form of a proposition: There does not exist
$\phi\in C^{0}(\mathbb{R}^{n})$ such that $\int_{\mathbb{R}^{n}}\phi(x)f(x)dx=f(0)$
for all $f\in C^{0}(\mathbb{R}^{n})$.
Proof: Prove by contradiction. Suppose the contrary that there exists
$\phi\in C^{0}(\mathbb{R}^{n})$ such that $\int_{\mathbb{R}^{n}}\phi(x)f(x)dx=f(0)$
for all $f\in C^{0}(\mathbb{R}^{n})$. To ease discussion, we only
consider the case that $n=1$. Firstly, we go to prove that $\phi(x)=0$
for all $x\in\mathbb{R}\setminus\{0\}$. Let $a\in\mathbb{R}\setminus\{0\}$
be arbitrary. Choose $\delta>0$ such that $0\notin(a-\delta,a+\delta)$.
Choose $g\in C^{0}(\mathbb{R})$ such that $g\geq0$, $g(a)=1$, and
$g=0$ outside $(a-\delta,a+\delta)$. Note that $g$ exists. For
example, we may choose $g$ to be
$$
g(x)=\begin{cases}
0, & \mbox{if }x\leq a-\delta\\
\frac{1}{\delta}(x-a)+1, & \mbox{if }x\in(a-\delta,a]\\
-\frac{1}{\delta}(x-a)+1, & \mbox{if }x\in(a,a+\delta)\\
0, & \mbox{if }x\geq a+\delta
\end{cases}.
$$
Note that, in particular, we have $g(0)=0$. Define $f=\phi g\in C^{0}(\mathbb{R})$.
By assumption, 
\begin{eqnarray*}
0 & = & f(0)\\
 & = & \int\phi(x)f(x)dx\\
 & = & \int\phi^{2}(x)g(x)dx,
\end{eqnarray*}
which implies that $\phi(a)=0$ (because if $\phi(a)\neq0$, then
$\phi^{2}g\geq0$, $\phi^{2}g$ is continuous, and $(\phi^{2}g)(a)>0$
$\Rightarrow\int\phi^{2}g>0$).
By continuity of $\phi$, it follows that $\phi(0)=\lim_{x\rightarrow0}\phi(x)=0$.
Hence, $\phi(x)=0$ for all $x\in\mathbb{R}$. Therefore, for any
$f\in C^{0}(\mathbb{R})$, $f(0)=\int\phi f=0$, which is obviously
a contradiction.
For the general case that $n>1$, the above proof continues to hold except that one needs to argue the existence of $g$. This can be done by invoking Urysohn Lemma in general topology.
