On the proof of the Fourier inversion theorem using (continuous) Fejér kernels

Question

Theorem (Fourier inversion theorem) Let $f \in L^1(\mathbb R^n)$, $\hat{f} := \mathcal{F}f \in L^1(\mathbb{R}^n)$. Then $$ (2 \pi)^{-\frac{n}{2}} \int_{\mathbb{R}^n} \hat{f}(\xi) e^{i \langle x, \xi \rangle} d \xi = \mathcal{F}^{-1}(\mathcal{F} f)(x) $$ holds almost everywhere.

Note: We defined the Fourier transform as $(\mathcal{F} f)(\xi) := \hat{f}(\xi) := (2 \pi)^{-\frac{n}{2}} \int_{\mathbb{R}^n} f(x) e^{-i \langle x, \xi \rangle} dx$.

I'll reproduce the proof from our lecture and add my questions after that.

Proof. We will only prove the statement for $n = 1$. For $\lambda > 0$ define the Fejér kernels $$ F_{\lambda}(x) := \frac{\lambda}{2 \pi} \int_{-1}^{1} (1 - | \theta |) e^{i x \theta \lambda} d\theta, $$ which we can also write as $$ F_{\lambda} := \lambda D_{\lambda^{-1}} F, \quad \text{where } F(x) = \frac{1}{2 \pi} \int_{-1}^{1} (1 - | t |) e^{i x t} dt $$ These kernels satisfy \begin{equation} \tag{1} \| f - f \ast F_{\lambda} \|_1 \xrightarrow{\lambda \to \infty} 0 \quad \text{and} \quad (f \ast F_{\lambda})(x) \to f(x) \ \text{a.e.} \end{equation} Using a simple substitution we obtain for $x \in \mathbb{R}$ \begin{align*} (f \ast F_{\lambda})(x) & = \int_{\mathbb{R}} f(t) \left[ \lambda \int_{-1}^{1} (1 - | \theta |) e^{i (x - t) \lambda \theta} d\theta\right] dt \\ & = \frac{1}{2 \pi} \int_{\mathbb{R}} f(t) \left[ \int_{-\lambda}^{\lambda} \left(1 - \frac{|u|}{\lambda}\right) e^{i u (x - t)} du \right] dt \\ & = \int_{-\lambda}^{\lambda} \left(1 - \frac{|u|}{\lambda}\right) \left[ \frac{1}{2 \pi} \int_{\mathbb{R}} f(t) e^{-i t u} dt \right] e^{i u x} du \\ & = \int_{-\lambda}^{\lambda} \left(1 - \frac{|u|}{\lambda}\right) \hat{f}(\theta) e^{i u x} du \\ & = \int_{| \theta | \le \sqrt{\lambda}} \left(1 - \frac{|u|}{\lambda}\right) \hat{f}(\theta) e^{i u x} du + \int_{\sqrt{\lambda} \le | \theta | \le \lambda} \left(1 - \frac{|u|}{\lambda}\right) \hat{f}(\theta) e^{i u x} du \\ & \xrightarrow{\lambda \to \infty} \int_{\mathbb R} \hat{f}(\theta) e^{i u x} du. \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \square \end{align*}

Questions

1. Is the theorem so much more difficult to prove for $n > 1$ or can the same techniques (e.g. multidimensional Fejér kernels if such a thing exists) with more involved calculations be applied?
2. The Wikipedia entry for Fejér kernels defines them differently, namely as sums (aka discrete integrals). Is the way we defined them "standard" way or rather unusual? What is the connection to the ones defined on Wikipedia?
3. Are the properties $(1)$ trivial or more involved to prove? If the latter is true, can you offer a hint on how to prove them?
4. Why can we explain the limit $\lambda \to \infty$ after splitting up the integral into the two integrals? Why couldn't we take the limit before and not split?

Answer 1

It will be convenient to define the function $$ \phi(\xi) = \begin{cases}1-|\xi|&\textrm{for}\,\,|\xi|<1,\\0&\textrm{for}\,\,|\xi|\geq1,\end{cases} $$ so that $$ F_\lambda(x)=\frac1{2\pi}\int_{\mathbb{R}}\phi(\xi/\lambda)e^{ix\xi}d\xi. $$

1. For $n>1$ we would have $$ F_\lambda(x)=\frac1{(2\pi)^n}\int_{\mathbb{R}^n}\phi(\xi/\lambda)e^{i\langle x,\xi\rangle}d\xi, $$ and the argument will stay the same.

2. What you defined is the standard Fejer kernel for the Fourier transform. The Wikipedia article focuses on the Fourier series, instead of the Fourier transform. The main idea in both cases is to use a triangular shaped function in the Fourier space.

3. One can calculate the Fejer kernel explicitly as $$ F_\lambda(x)=\frac1{2\pi}\int_0^\lambda(1-\xi/\lambda)(e^{ix\xi}+e^{-ix\xi})d\xi=\frac{1-\cos(\lambda x)}{\pi\lambda x^2}=\frac{2\sin^2(\lambda x/2)}{\pi\lambda x^2}, $$ which makes it clear that

   • $F_\lambda(x)\geq0$ for all $x$,
   • $F_\lambda(x)\to0$ as $\lambda\to\infty$ for $x\neq0$, and
   • $\int F_\lambda(x)dx=1$.

Then the standard argument involving approximation to the identity gives (1).

4. Perhaps you could justify it without the splitting. It is just a convenient choice to make the argument clear. With this splitting, it is obvious that the second integral goes to 0 as $\lambda\to\infty$. Then denote the first integral by $I_\lambda$, and note that $$ \big|I_\lambda-\int_{|u|<\sqrt{\lambda}}\hat f(u)e^{ixu}du\big| = \big|\int_{|u|<\sqrt{\lambda}}\frac{|u|}{\lambda}\hat f(u)e^{ixu}du\big|\leq\frac{1}{\sqrt\lambda}\|\hat f\|_{L^1}\to0, $$ as $\lambda\to\infty$. Without the splitting, we would not have the factor $\frac{1}{\sqrt\lambda}$ in front of the $L^1$-norm, because $u$ can be as large as $\lambda$.

Update: Let me elaborate on "the standard argument involving approximation to the identity."

First, suppose that $f$ is continuous and compactly supported. Then we have $$ f(x)=\int_{\mathbb{R}} f(x)F_\lambda(x-y)dy, $$ and so $$ f(x)-(f*F_\lambda)(x)=\int_{\mathbb{R}} \big( f(x)-f(y)\big)F_\lambda(x-y)dy. $$ Given any $\varepsilon>0$, we split the integral in right hand side into an integral over $|x-y|<\delta$ and one over $|x-y|>\delta$, with $\delta>0$ so small that $|f(x)-f(y)|<\varepsilon$ whenever $|x-y|<\delta$. This is possible because $f$ is uniformly continuous. Thus we have $$ \int_{|x-y|<\delta} | f(x)-f(y)|F_\lambda(x-y)dy\le \varepsilon\int_{|x-y|<\delta} F_\lambda(x-y)dy\le\varepsilon, $$ and $$ \int_{|x-y|\ge\delta} | f(x)-f(y)|F_\lambda(x-y)dy\le 2\|f\|_{L^\infty}\int_{|x-y|\ge\delta} F_\lambda(x-y)dy. $$ At this point, we can choose $\lambda$ large to make the integral arbitrarily small, thanks to the property $F_\lambda(t)\to0$ as $\lambda\to\infty$ whenever $t\neq0$. To conclude, $$ f*F_\lambda\to f\quad\textrm{uniformly.} $$

To proceed further, note that $$ \begin{split} \|f-f*F_\lambda\|_{L^1} &\leq\int\int |f(x)-f(y)|F_\lambda(x-y)dydx\\ &=\int\int |f(x)-f(x-t)|F_\lambda(t)dtdx, \end{split} $$ and as it should be familiar by now, split the inner integral into two pieces according to $|t|<\delta$ or $|t|>\delta$. The integral over $|t|<\delta$ is $$ \begin{split} \int_{\mathbb{R}}\int_{|t|<\delta}|f(x)-f(x-t)|F_\lambda(t)dtdx &=\int_{|t|<\delta}F_\lambda(t)\int_{\mathbb{R}}|f(x)-f(x-t)|dxdt\\ &\leq\sup_{|t|<\delta}\int_{\mathbb{R}}|f(x)-f(x-t)|dx, \end{split} $$ whose right hand side can be made arbitrarily small, by choosing $\delta>0$ small. Then the integral over $|t|>\delta$ is $$ \begin{split} \int_{\mathbb{R}}\int_{|t|>\delta}|f(x)-f(x-t)|F_\lambda(t)dtdx &=\int_{|t|>\delta}F_\lambda(t)\int_{\mathbb{R}}|f(x)-f(x-t)|dxdt\\ &\leq2\|f\|_{L^1}\int_{|t|>\delta}F_\lambda(t)dt, \end{split} $$ which again can be made arbitrarily small (for fixed $\delta>0$) by choosing $\lambda$ large. To conclude, $$ f*F_\lambda\to f\quad\textrm{in}\,\,L^1. $$

Now suppose that $f\in L^1$. Then for any given $\varepsilon>0$ there exists a continuous, compactly supported function $g$ such that $\|f-g\|_{L^1}<\varepsilon$. We have $$ \|f-f*F_\lambda\|_{L^1} \leq \|f-g\|_{L^1}+\|g-g*F_\lambda\|_{L^1}+\|g*F_\lambda-f*F_\lambda\|_{L^1}. $$ The first term is bounded by $\varepsilon$ by construction, and the second term can be made arbitrarily small by choosing $\lambda$ large. As for the third term, we have $$ \begin{split} \|g*F_\lambda-f*F_\lambda\|_{L^1} &\leq\int\int |g(y)-f(y)|F_\lambda(x-y)dydx\\ &=\int|g(y)-f(y)|\int F_\lambda(x-y)dxdy\\ &=\|g-f\|_{L^1}\leq\varepsilon. \end{split} $$ Thus, we conclude that $$ f*F_\lambda\to f\quad\textrm{in}\,\,L^1. $$

As for the almost everywhere convergence:

• First, as a warm-up, it is not difficult to extend the above argument to prove that $(f*F_\lambda)(x)\to f(x)$ whenever $f$ is continuous at $x$. But of course this is not enough, since a function can be discontinuous everywhere.
• Then by improving upon this argument, one can prove that $(f*F_\lambda)(x)\to f(x)$ whenever $x$ is a Lebesgue point of $f$. This gives what we want since almost every point is a Lebesgue point of $f$ if $f\in L^1$.

Let me know if you can reconstruct the proof. I think it would be instructive first to assume that $F_\lambda$ is supported in $|t|<1/\lambda$.