If $f(x) = e^{-|x|}$, show that $f''(x) - f(x) = -2\delta(x)$ (in the sense of distributions)

Question

Exercise :

If $f(x) = e^{-|x|}$, show that for its derivatives, it is : $f''(x)-f(x) = -2\delta(x)$, in the sense of distributions.

Attempt :

I am completely at a loss on how to handle such an exercise as we haven't studied distributions for more than 1-2 lessons, but I know that :

The function $f(x)$ is continuous and differentiable in $\mathbb R \setminus \{0\} = \mathbb R^*$, with derivative :

$$f'(x) = \begin{cases} -e^{-x}, & x>0 \\ e^x, & x<0\end{cases}$$

How would I proceed with modeling the derivatives now in the sense of distributions to prove the equation asked?

To be mentioned, the function $\delta(x)$ is the famous Dirac-delta function, such that:

$$\delta(x) = \begin{cases} +\infty & x=0, \\ 0 & x \neq 0 \end{cases}$$

Answer 1

$\delta$ is the distribution defined by $\langle \varphi,\delta \rangle = \varphi(0)$ for any test function $\varphi$.

Thus we want to show that for any test function $\varphi$, $$ \langle f''-f,\varphi \rangle = -2\varphi(0). $$ The derivative of a distribution is defined by $\langle f', \varphi \rangle = -\langle f,\varphi' \rangle $, mimicking the integration by parts formula. So we want to show that $$ \langle f,\varphi''-\varphi \rangle = -2\varphi(0). $$ Everything here is a function, so we can expand the pairing to its definition on functions, namely $$ \langle f,\varphi''-\varphi \rangle = \int_{-\infty}^{\infty} f(x)(\varphi''(x)-\varphi(x)) \, dx = \int_{-\infty}^0 e^{x} (\varphi''(x)-\varphi(x)) \, dx + \int_0^{\infty} e^{-x} (\varphi''(x)-\varphi(x)) \, dx. $$ Everything here is smooth on the intervals of integration, so we can integrate by parts on the first term in each: $$ \int_{-\infty}^0 e^{x} (\varphi''(x)-\varphi(x)) \, dx + \int_0^{\infty} e^{-x} (\varphi''(x)-\varphi(x)) \, dx = \left[ e^{x}\varphi'(x) \right]_{-\infty}^0 + \left[ e^{-x}\varphi'(x) \right]_0^{\infty} + \int_{-\infty}^0 e^{x} (-\varphi'(x)-\varphi(x)) \, dx + \int_0^{\infty} e^{-x} (\varphi'(x)-\varphi(x)) \, dx \\ = 0 + \int_{-\infty}^0 e^{x} (-\varphi'(x)-\varphi(x)) \, dx + \int_0^{\infty} e^{-x} (\varphi'(x)-\varphi(x)) \, dx, $$ and then integrating by parts again, $$ \int_{-\infty}^0 e^{x} (-\varphi'(x)-\varphi(x)) \, dx + \int_0^{\infty} e^{-x} (\varphi'(x)-\varphi(x)) \, dx = [-e^x\varphi(x)]_{-\infty}^0 + [e^{-x}\varphi(x)]_0^{\infty} + \int_{-\infty}^0 e^x(\varphi(x)-\varphi(x)) \, dx + \int_0^{\infty} e^{-x}(+\varphi(x)-\varphi(x)) \, dx \\ = -2\varphi(0), $$ as required.

Answer 2

Taking the Fourier transform (with the convention $\hat{f}(\xi)=\int f(x)e^{- ix\xi}dx$) of $f$ we get $$\hat{f}(\xi)=\frac{2}{1+\xi^2} $$ Hence $$\hat{f}+\xi^2\hat{f}=2 $$ The function $f$ can now be identified as a tempered distribution and the above equation still holds in this setting. Taking the inverse Fourier transform of the equation (in the sense of tempered distributions) we get $$f-f''=2\delta $$ where I used the fact that the (inverse) Fourier transform turns moltiplication by a power into differentiation and that the Fourier transform of $\delta$ is $1$ because for all $f\in \mathcal{S}(\mathbb{R})$ we have $$\hat{\delta} (f)=\delta (\hat{f})=\hat{f}(0)=\int 1\cdot f(x)dx=1(f) $$

Answer 3

Let $g\in\mathscr S$ be a test function. We have, by definition, $$ f[g]:=\int_{\mathbb R}g(x)\mathrm e^{-|x|}\mathrm dx\tag1 $$

Also, given $T\in\mathscr S^*$ an arbitrary distribution, we have, by definition, $$ T'[g]:=-T[g']\tag2 $$

Therefore, $$ (f''-f)[g]\overset{(2)}\equiv f[g'']-f[g]\overset{(1)}\equiv\int_{\mathbb R}(g''(x)-g(x))\mathrm e^{-|x|}\mathrm dx\tag3 $$

Now integrate by parts: $$ \begin{aligned} \int_{\mathbb R^+}(g''(x)-g(x))\mathrm e^{-x}\mathrm dx&\equiv\left[(g'(x)+g(x))\mathrm e^{-x}\right]_{\mathbb R^+}\\ \int_{\mathbb R^-}(g''(x)-g(x))\mathrm e^{+x}\mathrm dx&\equiv\left[(g'(x)-g(x))\mathrm e^{+x}\right]_{\mathbb R^+} \end{aligned}\tag4 $$

Adding both contributions, $$ (f''-f)[g]\overset{(4)}\equiv -2g(0)\tag5 $$

But this is precisely $-2\delta[g]$, as required.

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For a more direct, but slightly less rigorous, approach, you may as well just notice that $$ f(x)=\mathrm e^{+x}\Theta(-x)+\mathrm e^{-x}\Theta(+x)\tag6 $$ where $\Theta$ is the so-called step-function.

Using $\Theta'(x)=\delta(x)$, you get $$ f'(x)=\mathrm e^{+x}(\Theta(-x)-\delta(x))+\mathrm e^{-x}(\delta(x)-\Theta(+x))\tag7 $$ and $$ f''(x)=\mathrm e^{+x}(\Theta(-x)-2\delta(x)-\delta'(x))+\mathrm e^{-x}(-2\delta(x)+\Theta(+x)+\delta'(x))\tag8 $$

Adding both contributions, $$ f''(x)-f(x)=\mathrm e^{+x}(-2\delta(x)-\delta'(x))+\mathrm e^{-x}(-2\delta(x)+\delta'(x))\tag9 $$

This is precisely $-2\delta(x)$, as required. (Note that $\delta(x)f(x)=\delta(x)f(0)$ and $\delta'(x)f(x)=-\delta(x)f'(0)$).