# Fourier transform and convolution

Let $f \in L^1(\textbf{R})$ be such that $f'$ is continuous and $f' \in L^1(\textbf{R})$ . Find a function $g \in L^1(\textbf{R})$ such that

$$g(t) = \int_{-\infty}^{t}e^{u-t}g(u)\,du + f'(t)$$

Im pretty sure I should manipulate this with the fourier transform and that there is a convolution going on but I'm not really sure what to do now when the integral limit goes to $t$ instead of $\infty$.

• I'm not sure if that helps you, but $\int_{-\infty}^t e^{u-t} g(u)du = \left( e^{-t} \Bbb 1_{t\geq 0} \right) \ast g$ – Tryss Dec 11 '15 at 10:52

Using @Tryss hint, let $h(t)=\mathrm e^{-t}u(t)$, where $u(t)$ is the Heaviside step function. With this, $$(g*h)(t)\equiv\int_{-\infty}^{+\infty}g(u)\,\mathrm e^{-(t-u)}u(t-u)\,\mathrm du=\int_{-\infty}^t g(u)\,\mathrm e^{u-t}\,\mathrm du$$ where $*$ means convolution.

This means that your equation is $$g(t)=(g*h)(t)+f'(t)$$

Taking the Fourier transform, and using the Convolution Theorem, we find $$\hat g(\xi)=\hat h(\xi) \hat g(\xi)+2\pi i\xi \hat f(\xi)$$

Next, the Fourier Transform of $h(t)$ is easy: $$\hat h(\xi)=\frac{1}{1+2\pi i\xi}$$ which means that $$\hat g(\xi)\left[1-\frac{1}{1+2\pi i\xi}\right]=2\pi i\xi \hat f(\xi)$$

You can solve for $\hat g$, and take the Inverse Fourier Transform to find $g(t)$: $$\hat g(\xi)=(1+2\pi i\xi)\hat f(\xi)$$ and $$g(t)=f(t)+f'(t)$$

Explanation of $$(g*h)(t)\equiv\int_{-\infty}^{+\infty}g(u)\,\mathrm e^{-(t-u)}u(t-u)\,\mathrm du=\int_{-\infty}^t g(u)\,\mathrm e^{u-t}\,\mathrm du$$
My bad, I used the same letter twice: $u$ is both an integration variable and the Heaviside Step function. Im sorry for that. Lets rename the integration variable $w$ instead of $u$: $$(g*h)(t)\equiv\int_{-\infty}^{+\infty}g(w)\,\mathrm e^{-(t-w)}u(t-w)\,\mathrm dw$$ (this is the definition of the convolution)
Next, $u(x)$ is $1$ for positive $x$ and zero otherwise. This means that $u(t-w)$ is $1$ for $w<t$ and zero otherwise. In the integral above, where $w$ ranges from $-\infty$ to $+\infty$, the function $u(t-w)$ is zero for $w>t$ and $1$ for $w<t$, which means that the interval $w\in(t,\infty)$ doesnt contribute. This implies that we can change the integration interval from $(-\infty,+\infty)$ into $(-\infty,t)$, as the integrand is null from $(t+,\infty)$.
• $$(g*h)(t)\equiv\int_{\infty}^{+\infty}g(u)\,\mathrm e^{-(t-u)}u(t-u)\,\mathrm du=\int_{-\infty}^t g(u)\,\mathrm e^{u-t}\,\mathrm du$$ Can you explain this step? I know what the heaviside function is but I still can't manage to do this myself. The rest of the solution I understand :) – user269620 Dec 11 '15 at 13:21