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For $a>0$ and $x\ge 0$, split the convolution integral at $y=0$ and $y=x$. The three contributions are $e^{-ax}/(2a)$, $x{e}^{-ax}$ and $e^{-ax}/(2a)$. Evenness gives the displayed answer for all real $x$.

With Fourier transform convention $\tilde{f}(k)=\int f(x)e^{-ikx}dx$, one has $\tilde{f}(k)=2a/(a^2+k^2)$. The convolution theorem and Fourier inversion theorem therefore yieldThe value at $x=0$ is $\pi /(2a^3)$. Multiplying $f$ by $a/2$ instead gives a normalized Laplace distribution; its self-convolution is the density $a(1+a|x|)e^{-a|x|}/4$.

The triangular functionhas angular-frequency Fourier transform $4{\sin }^2(\xi /2)/{\xi }^2$.

Equivalently, this is the convolution of two unit-width rectangular functions: the overlap length is $1-|x|$ when $|x|<1$, and zero otherwise. Time rescaling and amplitude changes give triangular pulses with different widths and heights. The convolution theorem also obtains the displayed transform by squaring $\sin (\xi /2)/(\xi /2)$, with continuous value one at zero.

For integrable functions on the real line, equality of their Fourier transforms implies equality almost everywhere. To prove it, subtract and assume $\hat{h}=0$. Convolve $h$ with a Gaussian heat kernel $q_t$. The convolution theorem gives $\widehat{h\ast q_t}=0$, which is integrable, so the Fourier inversion theorem gives $h\ast q_t=0$ almost everywhere. Finally $q_t$ is an approximate identity: $\Vert h\ast q_t-h{\Vert }_1\le \int q_t(y)\Vert h(\cdot -y)-h{\Vert }_{1}dy\to 0$. This follows from translation continuity in Lp with exponent one and the vanishing Gaussian tails. Hence $h=0$ in $L^1$. If the two original functions are continuous, equality almost everywhere strengthens to equality everywhere, since a nonzero continuous difference has magnitude bounded below by a positive constant on some interval.

For an integrable linear response kernel $K$ and input $V$, the convolution theorem gives $\hat{I}(\omega )=\hat{K}(\omega )\hat{V}(\omega )$. The integrated output is therefore zero for every integrable input if and only if $\hat{K}(0)=\int K(t)dt=0$. For a sum of causal damped sines, $K(t)=\Theta (t){\sum }_j{a}_j{e}^{-{\gamma }_{j}t}\sin (p_{j}t)$, the condition is ${\sum }_j{a}_j{p}_j/({\gamma }_j^2+p_j^2)=0$. If all $a_j$ are positive, nonzero terms of both signs are needed; a term with $p_j=0$ vanishes identically.