Consider the following function defined for x≥0x\ge 0 real

f(x)={g(x)x

f(x) =

\begin{cases}

g(x) & x < x_0 \, ,\\
\infty & x= x_0 \, , \\
0 & x > x_0 \, ,

\end{cases}

wherein g(x)g(x) is a known function and x0≥0x_0 \ge 0.

In particular, if x0=0x_0=0, then f(x)f(x) may be represented as a kind of delta Dirac function.

I was wondering whether this piecewise function can be defined based on some known function. This will be useful for my further analysis.

Thank you.

Federiko

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1 Answer

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The object ff you aim to define is not a “function” f:R→Rf: \mathbb{R} \to \mathbb{R} in the usual sense, but actually a distribution, since ∞\infty is not a real number. Also the “delta function” is not a function in the strict sense, and it is actually defined as a distribution. The latter is an object that operates on functions ϕ:R→R\phi: \mathbb{R} \to \mathbb{R}, and for the delta function this is usually written as

δ[ϕ]=∫∞−∞δ(x)ϕ(x)dx=ϕ(0).\delta[\phi] = \int_{-\infty}^{\infty}\delta(x)\phi(x)dx = \phi(0)\,.

So in your case you could define ff as (if this is the kind of infinity you want at x0x_0):

f[ϕ]=∫∞0f(x)ϕ(x)dx=ϕ(x0)+∫x00g(x)ϕ(x)dx.f[\phi] = \int_{0}^{\infty}f(x)\phi(x)dx = \phi(x_0) + \int_{0}^{x_0}g(x)\phi(x)dx\,.

Here x≥0x\ge 0, why to you integrate from −∞-\infty? Thanks.

– Federiko

2 days ago

1

Sorry, that was a typo from copy-pasting the formula for δ\delta. The integral should start at 0.

– Xenos

2 days ago

Honestly I cannot view the meaning of ϕ\phi. Can we chose it arbitrary? or what? Thanks!

– Federiko

2 days ago

1

Yes, ϕ\phi is in some sense arbitrary (one usually takes it from some particular function space – smooth, compact support or the like). It is a so-called “test function” which is used to somehow “probe” the distribution. By allowing all possible test functions and specifying how the distribution ff (or δ\delta) behaves if it is evaluated on each of them (as in the formulas in my answer), you specify the distribution itself.

– Xenos

2 days ago