Continuity of a function in $H^s(\mathbb R^n)$, for sufficiently large $s.$ The following is a proposition and its proof regarding the Sobolev embedding theorem from the textbook on Partial Differential Equations. (Taylor, M. Partial Differential Equations, Vols. 1-3, Applied Math. Sciences, 115-117, Springer-Verlag, New York, 1996.)

Proposition 1.3. If $s > n/2$, then each $u ∈ H^s(\mathbb{R}^n)$ is bounded and
continuous.


Proof. By the Fourier inversion formula, it suffices to prove that $\hat{u}(ξ)$
belongs to $L^1(\mathbb{R}^n)$. Indeed, using Cauchy’s inequality, we get
$$ \int |\hat{u}(ξ)| dξ =\int|\hat{u}(ξ)|(\xi)^s(\xi)^{-s}\leq \left(\int |\hat{u}(\xi)|^2(\xi)^{2s}\right)^{\frac{1}{2}}\left(\int (\xi)^{-2s}\right)^{\frac{1}{2}}$$
Since the last integral on the right is finite precisely for s > n/2, this
completes the proof.

Even though I understand the construction of the proof, I want to clarify why $\hat{u}(\xi)\in L^1(\mathbb{R}^n)$ is sufficient to show that each $u ∈ H^s(\mathbb{R}^n)$ is bounded and
continuous. I understand that this comes from $L^1$ Theory of the Fourier Transform, which states that
$$f\in L^1(\mathbb{R})\rightarrow \hat{f} \in C_b^0(\mathbb{R}^n)$$
so $\mathcal{F}$ is a linear, bounded, continuous transformation. (but not invertable) However, if $\hat{u}(\xi)\in L^1(\mathbb{R}^n)$, and we already have $u(x)\in L^1(\mathbb{R}^n)$ then we can use the Inversion theorem that says that $u$ is continuous almost everywhere. (but not everywhere?)
I believe this may be adressed in this post Property of inverse of Fourier transform
It would be helpful if someone could please clarify my though process and/or point out the error in my justification.
I also noticed that $s>n/2$ implies $H^s(\mathbb{R}^n)$ is a Banach space, however I don't see this property being used in the proof.
 A: Fourier transform has many different (but equivalent) definitions. I'll take the following version:
$$\hat f(\xi)=\int_{\mathbb{R}^n} f(x)e^{-2\pi ix\cdot\xi}\,dx,\qquad f^\vee(x)=\int_{\mathbb{R}^n} f(\xi)e^{2\pi ix\cdot\xi}\,d\xi.$$
Now let me explain the logic of “$u$ is bounded and continuous”. This kind of situation often happens in analysis. In many circumstances, we say that a function $u$ has some propeties if there exists another function $v$ satisfies the same properties and $u=v$ almost everywhere.
Since $g(\xi):=\hat{u}(\xi)\in L^1(\mathbb{R}^n)$, we can define a function
$$v(x)=\int_{\mathbb{R}^n} g(\xi)e^{2\pi ix\cdot\xi}\,d\xi=\int_{\mathbb{R}^n} \hat{u}(\xi)e^{2\pi ix\cdot\xi}\,d\xi,$$
then $v(x)=\hat g(-x)$. Since $g\in L^1(\mathbb{R}^n)$, we know that $\hat g\in C_b^0(\mathbb{R}^n)$, hence $v\in C_b^0(\mathbb{R}^n)$.
Now, if $u\in L^1(\mathbb{R}^n)$, by the Inversion theorem, we have
$$u(x)=\int_{\mathbb{R}^n} \hat{u}(\xi)e^{2\pi ix\cdot\xi}\,d\xi,\qquad \text{a.e. }x\in\mathbb R^n.\tag{1}$$
Thus, $u=v$ almost everywhere, so we can view $u$ and $v$ as the same function. Since $v$ is bounded and continuous, we can say that $u$ is bounded and continuous.
However, here we don't know whether $u$ is in $L^1(\mathbb{R}^n)$ or not. We only have that $u\in H^s(\mathbb{R}^n)\subset L^2(\mathbb{R}^n)$. We prove that in this case, $(1)$ also holds.

Lemma. If $u\in L^2(\mathbb{R}^n)$ and $\hat u\in L^1(\mathbb{R}^n)$, then
$$u(x)=v(x):=\int_{\mathbb{R}^n} \hat{u}(\xi)e^{2\pi ix\cdot\xi}\,d\xi,\qquad \text{a.e. }x\in\mathbb R^n.$$

Proof of Lemma. For any $\phi\in C_c^\infty(\mathbb R^n)$, we prove that
$$\int_{\mathbb{R}^n}u(x)\overline{\phi(x)}\,dx=\int_{\mathbb{R}^n}v(x)\overline{\phi(x)}\,dx.\tag{2}$$
Indeed, we have
\begin{align*}
\int_{\mathbb{R}^n}u(x)\overline{\phi(x)}\,dx&=\int_{\mathbb{R}^n}\hat u(\xi)\overline{\hat \phi(\xi)}\,d\xi\\
&=\int_{\mathbb{R}^n}\hat u(\xi)\overline{\int_{\mathbb{R}^n}\phi(x)e^{-2\pi ix\cdot\xi}\,dx}\,d\xi\\
&=\int_{\mathbb{R}^n}\hat u(\xi)\int_{\mathbb{R}^n}\overline{\phi(x)}e^{2\pi ix\cdot\xi}\,dx\,d\xi\\
&=\int_{\mathbb{R}^n}\overline{\phi(x)}\int_{\mathbb{R}^n}\hat u(\xi)e^{2\pi ix\cdot\xi}\,d\xi\,dx\\
&=\int_{\mathbb{R}^n}\overline{\phi(x)}v(x)\,dx.
\end{align*}
Hence, $(2)$ holds for all $\phi\in C_c^\infty(\mathbb R^n)$. Now it is a routine exercise of real analysis to check that $u=v$ almost everywhere.
