# How to prove that $\lim\limits_{\epsilon\rightarrow 0}\int_{-\infty}^{\infty}f_\epsilon(x)g(x)dx=g(0)$ (Dirac delta function))

I'm currently studying the Dirac delta function using a textbook which unfortunately provides only partial solutions to its explanations. Why does $$\lim\limits_{\epsilon\rightarrow 0}\int_{-\infty}^{\infty}f_\epsilon(x)g(x)dx=g(0)$$ if $$f_\epsilon=\frac{1}{2\epsilon}e^{-\frac{|x|}{\epsilon}}$$? Any help is very much appreciated.

• @ViktorGlombik What makes you think so? – lisyarus Apr 18 at 11:05
• Nevermind, I thought of something else. – Viktor Glombik Apr 18 at 11:07
• What is $g$ here? The answer depends on the type of $g$ you want to consider. – Kabo Murphy Apr 18 at 11:28

Use a change of variable: $$\int_{-\infty}^{\infty} \frac{1}{2\epsilon} \, e^{-|x|/\epsilon} g(x) \, dx = \{ y = x/\epsilon \} = \frac{1}{2} \int_{-\infty}^{\infty} e^{-|y|} \, g(\epsilon y) \, dy \\ \to \frac{1}{2} \int_{-\infty}^{\infty} e^{-|y|} \, g(0) \, dy = \frac{1}{2} \left( \int_{-\infty}^{\infty} e^{-|y|} \, dy \right) \, g(0) = g(0)$$ The dominated convergence theorem has been used when taking the limit.
Set $$f(x) = \frac{1}{2}e^{-|x|}$$, so $$\frac{1}{\epsilon} f(x/\epsilon) =f_\epsilon$$ and $$||f||_{L^1} = 1$$ for $$\epsilon\in (0,1]$$, so for any test functions $$g\in C_c^\infty$$ you have $$\int\frac{1}{\epsilon} f(x/\epsilon)g(x)dx = \int f(x)g(\epsilon x)dx$$ $$=\int f(x)[g(0) + g'(0)\epsilon x + ...]dx$$ $$=g(0)\int f(x)dx + \int_{}f(x)[\epsilon xg'(0) + \frac{g''(0)(x\epsilon)^2}{2}+ ...]dx$$ The expansion is valid by the support and regularity assumption and $$f\in \mathcal{S}$$. So you have the limit $$\int\frac{1}{\epsilon} f(x/\epsilon)g(x)dx \rightarrow g(0)\int f(x)dx = g(0)$$ hence $$f_\epsilon \rightarrow \delta$$ in the distributional sense