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Let $f \in L^1(\mathbb{R})$ be such that $f'$ is continuous and $f' \in L^1(\mathbb{R})$. Find a function $g \in L^1(\mathbb{R})$ such that $$g(t) = \int_{-\infty}^{t} e^{u-t} g(u) du - f'(t), \quad t\in \mathbb{R}$$

We identify the integral as the convolution $h*g$ with $h(x) = e^{-x}(1-\theta(x))$. Fourier transforming both sides yields $$\hat{g}(\omega) = \hat{h}(\omega)\hat{g}(\omega)-i\omega f(\omega)$$ and so we seek to find $\hat{h}$. The definition of the Fourier transform my book uses would lead us to $$\hat{h}(\omega) = \int_{\mathbb{R}} e^{-x}(1-\theta(x))e^{-i\omega x} dx $$ which clearly is divergent. How can I salvage this?

Lozansky
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1 Answers1

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Maybe assume that you have more smoothness in $f$ and then take the derivative of both sides of the equation. If $f$ has a second derivative, then \begin{align*} g'(t) &= g(t) - \int_{-\infty}^t e^{u-t}g(u) d u - f''(t) \\ &=g(t) - (g(t) + f'(t)) - f''(t) \\ &= -f'(t) -f''(t). \end{align*} Thus we get that $g(t) = C -f(t)-f'(t)$, and we choose $C=0$ to get $g\in L^1(\mathbb{R})$. So if $f$ has a second derivative, $g$ (if it exists) must be $-f -f'$. We can use this result to make an educated guess about the general solution.

Here's the rest of my answer if you don't care about working it out:

We work off the assumption that this holds for $f$ even without a second derivative. We wish to show that \begin{equation*} -f(t) - f'(t) = - \int_{-\infty}^t e^{u-t}(f(u) + f'(u)) d u - f'(t) \end{equation*} holds. This is equivalent to showing $$ f(t) = \int_{-\infty}^t e^{u-t}(f(u) + f'(u)) d u.$$ We can evaluate the integral directly: \begin{align*} \int_{-\infty}^t e^{u-t}(f(u) + f'(u)) du &= e^{-t} \int_{-\infty}^t e^{u}(f(u) + f'(u)) du \\ &= e^{-t} \int_{-\infty}^t \frac{d}{du} [e^u f(u)]du \\ &=e^{-t} (e^tf(t)) \\ &= f(t). \end{align*} Hence, $g(t) = -f(t) - f'(t)$ is a solution.

Trevor Norton
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