If $f\in L^2(\mathbb{R^d})$ and $(\partial f / \partial x)^{\alpha} \in L^2$ for all $|\alpha|\leq n$ then f then f is continuous and bounded

Question

Hi I found this exercise in Stein's measure theory book and was wondering if somebody could help me out by giving a simple answer (ie no PDE's theory).

Let n be the smallest integer bigger than $\frac{d}{2}$.If $f\in L^2(\mathbb{R^d})$ and $(\partial f / \partial x)^{\alpha} \in L^2$ in the weak sense for all $|\alpha|\leq n$ then f can be modified on a set of measure zero so that f is continuous and bounded.

The hint given is to write f in terms of its fourier transform $F(f)$ , and show that $F(f) ∈ L^1(\mathbb R^d)$ by the Cauchy-Schwarz inequality.

Now I know from a previous exercise the following fact:

If $f∈L^2(R^d)$ and $g= (\partial f / \partial x)^{\alpha}\in L^2(\mathbb R^d)$ Then $(2πiξ)^α F(f)(ξ) = F(g)(ξ) ∈ L^2(\mathbb{R^d})$

I think using this should lead me somewhere but I tried everything and am stuck..tried dotting f with the indicator but the indicator is not integrable, also tried dotting f with ξ and using cauchy shwartz and again to no avail.

Could someone help me please? Also why would the fourier transform being integrable imply f is bounded and continuous almost everywhere? Thanks in advance for any help.

Answer 1

Observe, we have that \begin{align} \int_{\mathbb{R}^d}|\mathcal{F}(f)(\xi)|\ d\xi \leq&\ \left(\int_{\mathbb{R}^d}|(1+|\xi|^2)^{\alpha/2}\mathcal{F}(f)(\xi)|^2\ d\xi\right)^{1/2} \left(\int_{\mathbb{R}^d}\frac{d\xi}{(1+|\xi|^2)^\alpha} \right)^{1/2} \\ \leq&\ C\left(\bigg\|\frac{\partial^\alpha}{\partial x^\alpha}f\bigg\|_{L^2}+\|f\|_{L^2}\right) <\infty. \end{align} Note we have used the fact that $\alpha> \frac{d}{2}$. Hence $\mathcal{F}(f) \in L^1(\mathbb{R}^d)$. In particular, by the Fourier inversion formula, we see that \begin{align} f(x) = \int_{\mathbb{R}^d}\mathcal{F}(f)e^{2\pi i\xi\cdot x}\ d\xi \end{align} pointwise almost everywhere. Since Fourier transform of an $L^1$ function $g$ is continuous and $\lim_{x \rightarrow \infty} \mathcal{F}(g)(x)=0$, then it follows that $f$ is continuous and bounded since $f$ is the Fourier transform of an $L^1$ function, namely $\hat f(\xi)$.

Edit: Observe that \begin{align} \int_{\mathbb{R}^d}|(1+|\xi|^2)^{\alpha/2}\mathcal{F}(f)(\xi)|^2 \ \leq& C\int_{\mathbb{R}^d}(1+|\xi|^{2\alpha})|\mathcal{F}(f)(\xi)|^2\ d\xi\\ \leq& C \int_{\mathbb{R}^d}|\mathcal{F}(f)(\xi)|^2\ d\xi + C\int_{\mathbb{R}^d}||\xi|^\alpha\mathcal{F}(f)(\xi)|^2\ d\xi\\ =& C\| f\|^2_{L^2}+C\bigg\|\frac{\partial^\alpha}{\partial x^\alpha}f\bigg\|_{L^2}^2 \end{align} where the last equality follows from Plancherel's theorem.

Answer 2

It is from Stein's Real Analysis, chapter 5, exercise 16.

I prefer $(1+|\xi|^n)$.

(1) $\left(\frac{\partial}{\partial x_k}\right)^n f\in L^2(\mathbb R^d)\implies (\xi_k)^n \hat f(\xi)\in L^2(\mathbb R^d)$.

(2) $|\xi|\leq \sqrt{d}\max\{|\xi_1|,\ldots, |\xi_n|\}\implies |\xi|^n\leq d^{\frac{n}{2}}\cdot \max\{|\xi_1|^n,\ldots, |\xi_n|^n\}\leq d^{\frac{n}{2}}\cdot(\sum_{k=1}^d |\xi_k|^n)$.

(3) With $g(\xi)=(1+|\xi|^n)\cdot \hat f(\xi)$, we have $$|g(\xi)|\leq (1+d^{\frac{n}{2}}\sum_{k=1}^d |\xi_k|^n)\cdot |\hat f(\xi)|\implies g(\xi)\in L^2(\mathbb R^d).$$ Let $h(\xi)=\frac{1}{1+|\xi|^n}$, and $n>\frac{d}{2}$ implies $h(\xi)\in L^2(\mathbb R^d)$. $$\int |\hat f(\xi)| d\xi=\int |h(\xi) g(\xi)|\leq \Vert h\Vert_{L^2(\mathbb R^d)}\cdot \Vert g\Vert_{L^2(\mathbb R^d)}.$$