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If the constant symmetric matrix $A$ satisfies $A\xi \cdot \xi \ge \theta |\xi {|}^2$, then every smooth function $u$ with compact support satisfiesIndeed, the Fourier transform of a derivative and the Plancherel theorem turn the squared norms into integrals with Fourier multipliers $|\xi {|}^4$ and $(A\xi \cdot \xi {)}^2$.

A Fourier multiplier operator $T_m$ is defined by $\widehat{T_{m}f}(\xi )=m(\xi )\hat{f}(\xi )$. On $L^2$, a bounded symbol gives a bounded operator by the Plancherel theorem, while an unbounded symbol defines a closed linear operator on the functions for which $m\hat{f}$ remains square-integrable.

Functions with pairwise disjoint, or uniformly finitely overlapping, Fourier supports are almost orthogonal in $L^2$ by the Plancherel theorem.

For $\mathcal{H}f(x)={\pi }^{-1}\mathrm{pv}\int f(y)/(x-y)dy$ and $\hat{f}(\xi )=\int e^{-ix\xi }f(x)dx$, the Fourier transform multiplier is $-i\mathrm{sgn}\xi$. The Heaviside function transform and the double-transform reflection identity give the sign. Plancherel theorem proves the $L^2$ isometry, and multiplication by $i\xi$ proves commutation with differentiation. The map need not preserve the Schwartz space.

If a finite measure on ${\mathbb{R}}^n$ has square-integrable Fourier transform, the Plancherel theorem gives an $L^2$ inverse transform $g$. Pairing with every Schwartz function and using Fourier inversion identifies the measure with $gdx$. This establishes absolute continuity, rather than presuming it. For a positive measure, $g\ge 0$ and $g\in L^1$ as well. Multiplication by $f\in L^{\infty }(\mu )$ yields $\Vert \widehat{fd\mu }{\Vert }_2\le \Vert \hat{\mu }{\Vert }_2\Vert f{\Vert }_{L^{\infty }(\mu )}$.

Use $\hat{f}(\xi )=\int f(x)e^{-ix\xi }dx$ and let $P_j$ be the orthogonal projection onto the closed span of the orthonormal translates and dilates of a scaling function. If $\hat{f}$ is supported in $[-R,R]$ and $R<\pi 2^j$, thenIndeed, the Plancherel theorem writes the coefficient against ${\phi }_{j,k}$ as $2^{-j/2}(2\pi {)}^{-1}\int \hat{f}(\xi )\overline{\hat{\phi }(2^{-j}\xi )}{e}^{ik{2}^{-j}\xi }d\xi$. With $\xi =2^j\theta$, these are $2^{j/2}$ times the Fourier series coefficients of $\hat{f}(2^j\theta )\overline{\hat{\phi }(\theta )}$ on $[-\pi ,\pi ]$. The Parseval identity proves the formula. It shows that continuity and unit modulus at zero imply density of the refinement spaces, and conversely that density forces this unit modulus when the Fourier transform is continuous at zero.

With $\hat{\varphi }(\xi )=\int \varphi (x)e^{-ix\xi }dx$ in the Plancherel theorem sense, the integer translates of $\varphi \in L^2(\mathbb{R})$ are orthonormal exactly whenThe Fourier coefficients of the periodized energy are the translate inner products. Uniqueness of Fourier coefficients in L1 therefore proves both directions. No absolute integrability assumption on $\varphi$ is required.