# Delta Dirac Function

I realize it's not a function in the classical sense, but how would one show that the delta dirac function is a distribution i.e. how do I show it's continuous and linear given that it's not truly a function?

What is your definition of the Dirac delta function??

Given any x_0 ∈ Rn, the delta function is the distribution, δ_{x_0} :D(Rn)→C
given by the evaluation of a test function at x_0: ⟨δ_{x_0} , φ⟩ = φ(x_0)

The set of set of test functions is a vector space. The reals are are also a vector space. Use that for linearity.
Test functions are smooth, use that for continuity.

Hi sdickey9480!

I think your question really amounts to 'what is a distribution'? As mentioned by the previous posters it has to do with test functions, and more generally with special vector spaces (usually complete ones and those endowed with a norm).

An amazing triumph of functional analysis is representing vectors (in this case non pathological functions) in terms of their actions on other vectors. By action on other vectors, I mean given any vector $v$ in the vector space $V$, define a mapping $\hat{v}:V\rightarrow \mathbb{R}$. This mapping is given by the Riesz Representation Theorem, and in our case it means $\hat{v}(g):=\int fg \mathrm{d}x$

$\hat{v}$ is called linear because $\hat{v}(f+g)=\hat{v}(f)+\hat{v}(g)$.

The second important property we want $\hat{v}$ to have is that of continuity. Another surprising result of functional analysis says that a functional (any linear map from the vector space into the reals, like $\hat{v}$ for example) is continuous if and only if it is bounded in the operator sense. That is, $\hat{v}$ is bounded if and only if sup$\{\hat{v}(f): ||f||_\infty = 1 \}< \infty$

What I have done is built the necessary machinery to generalize functions. What I have shown is that any vector (or in this case non-pathological function) can be thought of as a continuous linear functional. A distribution is then just one of these continuous linear functionals.

So to answer your question, the dirac delta function $\delta$ is defined as a functional, mapping some space of functions to the real line by $\delta (f) = \int f\delta \mathrm{d}x := f(0)$. It is clear that $\delta$ is linear because integral is linear (actually strictly speaking the integral doesn't make sense, hence the need for generalized functions to begin with. We really define $\delta$ to be linear).

Why is $\delta$ continuous? Because if $||f||_\infty = 1$ and $f$ is continuous, then $f(x)\leq 1$ for any $x$. Hence $\delta$ is bounded by $1$, and therefore continuous.