Function that is non-zero only at one point. I am searching for, if there exists, a continuous function $f(x)$ such that $f(x) = 0$ for all values of $x$, with the exception of one point (say $\tilde x$) where $f(\tilde x)\neq0$.
This is possible if and only if $\tilde x$ is an isolated point of the domain.

Example 1: For example, the domain could be $$(-\infty,\tilde x - \epsilon)\cup \{\tilde x\}\cup (\tilde x +\epsilon,\infty)$$ if it is a subset of $\mathbb R$ (which, incidentally, is not specified) and the function rule could be
$$f(x) = \begin{cases}
0 &amp; \textrm{ if }x &lt;\tilde x - \epsilon\\
c &amp; \textrm{ if }x = \tilde x \\
0 &amp; \textrm{ if }x &gt; \tilde x +\epsilon
\end{cases}
$$

where $\epsilon$ is a positive constant and $c$ is a nonzero constant.

For example, you could have $$f:(-\infty,-1)\cup \{0\}\cup (1,\infty) \rightarrow \mathbb R$$

with

$$f(x) = \begin{cases}
0 &amp; \textrm{ if }x &lt;-1\\
1 &amp; \textrm{ if }x = 0 \\
0 &amp; \textrm{ if }x &gt;1
\end{cases}
$$

Example 2: An even simpler example of such a function would be
$$f:\{0,1\} \rightarrow \mathbb R$$
with
$$f(x) = \begin{cases}
1 &amp; \textrm{ if }x = 0 \\
0 &amp; \textrm{ if }x =1
\end{cases}
$$
Everyone seems to be mistaken about what continuity of a function actually means. The Dirac Delta function answers the question (as others have already suggested). The Dirac delta is a continuous function; as defined by its integrability i.e. the n-dimensional integral of any n-dimensional Dirac delta function is unity (it would be naturally non-integrable if it was discontinuous).

Looking at it from a slightly more intuitive angle; it can be recognised that the Dirac delta is derived via the integration of a complex valued exponential function (which are naturally continuous).