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Take the Fourier transform in $x$. Since $\widetilde{{\theta }_{xx}}=-k^2\tilde{\theta }$, the transformed heat equation isThereforeThe given transform pair and the convolution theorem yield the heat-kernel solution

The convolution isWe claim that the $n$-fold convolution isThis is true for $n=1$. If it holds for $n-1$ and $x\ge 0$, thenand the convolution vanishes for $x<0$. This is the gamma density from repeated exponential convolution.

The Fourier transform is

The convolution theorem statesIndeed, Fubini and $u=x-y$ giveSince $\hat{F}(k)=(1+ik{)}^{-1}$, induction immediately verifies

For Parseval identity, take $g(x)=\overline{f(-x)}$. ThenUsing Fourier inversion at zero, which follows from the supplied delta identity,and the convolution theorem gives

Apply this to $F_{n+1}$. Sinceone obtains the Even rational Parseval integral

Insert the transform definitions and use Fubini:For the indicator of $(-1,1)$,Putting $g=f$ and $x=0$ in the convolution theorem givesand hence

Use normalized convolution and Fourier coefficients, and define the large spectrumBy the Parseval identity and $0\le f\le 1$,so $|\Gamma |\le {ϵ}^{-2}{\delta }^{-1}$.

The convolution theorem givesIf $y\in B(\Gamma ,ϵ)$, the part over $\Gamma$ is at mostbecause $|\hat{f}(\gamma )|\le \delta$. On the complementary spectrum, $|\hat{f}(\gamma )|<ϵ\delta$ and $|\gamma (y)-1|\le 2$, so the contribution is less than $2ϵ{\delta }^2$. The required difference is therefore less than $3ϵ{\delta }^2$.

With the convention in convolution theorem, absolute integrability permits Fubini:

If $g\in L^2$ is orthogonal to every translate of $f$, then the correlation $g\ast \tilde{f}$ vanishes. By convolution theorem,
$\hat{g}\overline{\hat{f}}=0$ almost everywhere. Since $\hat{f}$ vanishes only at one point, a null set, $\hat{g}=0$ almost everywhere. By Plancherel theorem, $g=0$. Orthogonal-complement characterization proves density in $L^2$.