Let $f \in L^1(\mathbb{R}^n)$ with $\hat{f} \in L^1(\mathbb{R}^n)$. Then for almost every $x$: $f(x) = \int_{\mathbb{R}^n} \hat{f}(\xi)e^{2\pi i x \cdot \xi}\,d\xi$

If additionally $f$ is continuous, this holds for all $x$.

The Fourier transform extends uniquely from $L^1 \cap L^2$ to a unitary operator on $L^2(\mathbb{R}^n)$: $\mathcal{F}: L^2(\mathbb{R}^n) \to L^2(\mathbb{R}^n)$ satisfying $\|\mathcal{F}[f]\|_{L^2} = \|f\|_{L^2}$ and $\mathcal{F}^{-1} = \mathcal{F}^*$ (adjoint).

Moreover, for $f, g \in L^2$: $\int_{\mathbb{R}^n} f(x)\overline{g(x)}\,dx = \int_{\mathbb{R}^n} \hat{f}(\xi)\overline{\hat{g}(\xi)}\,d\xi$

Let $f$ be piecewise smooth and $2\pi$-periodic. Then the Fourier series converges pointwise to: $\frac{1}{2}[f(x^+) + f(x^-)]$ where $f(x^+)$ and $f(x^-)$ are right and left limits.

If $f$ is continuous at $x$, the series converges to $f(x)$.

For $f \in H^s(\mathbb{R}^n)$ (Sobolev space with $s$ derivatives in $L^2$), defined by: $\|f\|_{H^s}^2 = \int_{\mathbb{R}^n} (1 + |\xi|^2)^s|\hat{f}(\xi)|^2\,d\xi < \infty$

If $s > n/2$, then $f \in C^0(\mathbb{R}^n) \cap L^\infty$ and: $|f(x)| \leq C\|f\|_{H^s}$

This explains why solutions with sufficient regularity in Sobolev spaces are automatically continuous.