Consider the following problem. Is the hypothesis that $f\in C^\infty$ necessary, or could we weaken it and assume just that $f$ is continuous?
Let $\hat f$ denote the Fourier transform of the function $f$ on $\mathbb R$. Suppose $f\in C^\infty$ and $\|\hat f(\xi) \|_2 \le \alpha$, $ \|\xi^{1+\epsilon} \hat f\|_2 \le \beta$, for some $\epsilon >0$. Find a bound on $\| f(x)\|_\infty$ in terms of $\alpha$, $\beta$, and $\epsilon$.
Proposed Solution (assuming only that $f$ is continuous): It suffices to bound $\|\hat f\|_1$ in terms of $\alpha$, $\beta$, $\epsilon$. Once we know $\hat f \in L_1$, we can write
$$F^{-1}\hat f(\xi) = \int_{\mathbb R} \hat f(\xi)e^{2\pi i\xi}\ d\xi,$$
since the $L^2$ Fourier inverse definition and $L^1$ Fourier inverse definition agree on functions in $L^1\cap L^2$. Since $F^{-1}\hat f(\xi)$ is continuous, and $f(x)$ is the unique member of that $L^2$ equivalence class that is continuous, we have
$$f(x)= \int_{\mathbb R} \hat f(\xi)e^{2\pi i\xi}\ d\xi.$$
Then our original bound on $\| \hat f\|_1$ gives a bound on $\|f\|_\infty$.
For the bound on $\|\hat f\|_1$, split $\hat f$ as $\hat f=\hat f\chi_{[-1,1]} + f\chi_{[-1,1]^c}$. On the piece $[-1,1]$, we note $[-1,1]$ is a finite measure space, so the $L^1$ norm is bounded by a constant times the $L^2$ norm (by the usual Cauchy-Schwarz argument). On $[-1,1]^c$, write $\hat f = \xi^{-(1+\epsilon)}(\xi^{1+\epsilon}\hat f)$ and again apply Cauchy-Schwarz.