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Let $\mathcal{F}$ be the fourier transform operator and let $T_R$ = $\mathcal{F}^* \chi_R\mathcal{F}$ where $\chi_R$ is the indicator function on the ball of radius $R$. Hence $T_R$ is the fourier multiplier operator with the indicator function on ball of radius R. I am interested in understanding the question for what $p$ with $1\leq p\leq \infty$ does $T_R(f) \to f$ in $L^p$ as $R\to \infty$ for all $f\in L^p(\mathbb{R}^n)$ ?

I am only interested in $n\geq2$.

Now first I am a little confused about how to make the question precise. Now $T_R(f) = f*\hat{\chi_R}$ and by stationary phase arguments I know that the exact asymptotics of $\hat{\chi_R}(\xi)$ is like $|\xi|^{-(n+1)/2}$ as $\xi \to \infty$ and hence $\hat{\chi_R} \in L^p(\mathbb{R}^n)$ iff $p> \frac{2n}{n+1}$. Hence the problem definitely doesn't make sense (i.e. there is trivial schwartz function for which the statement fails) for $1\leq p \leq \frac{2n}{n+1}$. By using Young's inequality, we know that $T_R(f)$ is in $L^1_{loc}$ for $ f \in L^p$ for the range $ \frac{2n}{n+1}<p < \frac{2n}{n-1}$ hence the problem is potentially meaningful/interesting in this range. But what about $p \geq \frac{2n}{n-1}$? Is there a clear counterexample or an argument which rules out this range of $p$? I am not even sure whether $T_R(f)$ is in $L^1_{loc}$ for $p$ in this range.

In a lot of places I have read that the original problem is equivalent to the problem of boundedness of the operator $T_R$ from $L^p \to L^p$ (Stein mensions this in his book Harmonic Analysis page 389). I understand that if $T_R$ is bounded, then the original problem is solved for that $p$ by a simple density argument. However I do not see why we really need to have boundedness of the operators $T_R$. Grafakos in his book Modern Fourier Analysis "proves" that we really need boundedness of $T_R$ in the exercise 10.2.1 page 366 and the proof is via uniform boundedness principle. However I do not see how we can apply uniform boundedness principle as the principle only applies if we apriori know that the individual operators are bounded.

For example consider an infinite dimensional Banach space $X$. By axiom of choice, let us chose a hamel basis and collect an countable infinite among them $x_1, x_2,..$. We define the unbounded operators $T_n:X \to X$ by defining them on the chosen basis vectors and extending linearly. Define $T_n(x_i) = i*x_i$ for $i\geq n$ and zero for $i<n$, and zero on all other basis vectors. We clearly see that $T_n$ are unbounded operators, for every $f\in X, sup\|T_n(f)\| < \infty$ and $T_n(f) \to 0$ for every $f\in X$.

So to summarize I have 2 questions:

1) Whether the question $T_R(f) \to f$ in $L^p$ for all $f\in L^p$ makes sense (or does not makes sense) for $p \geq \frac{2n}{n-1}$

2) How does $T_R(f) \to f$ in $L^p$ for all $f\in L^p$ imply the boundedness of the opertors $T_R$

I am sorry for the long post. Any help would be appreciated!

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You can try to simplify the problem, considering instead the equivalence of the $L^p$-boundedness of the operator $T_1$ and the convergence $T_Rf\to f$ in $L^p$ for every $f\in L^p$.

In the case when $T_1$ is bounded in $L^p$, you can find a prove in the book of Grafakos, theorem 10.2.4. Recall the identity $T_Rf(x)=T_1f(\cdot/R)(Rx)=\text{Dil}_RT_1\text{Dil}_{R^{-1}}f$. For the converse, I never had thought about the issue you pointed, but I think you can consider the sequence of bounded operators $\phi T_R\phi$, where $\phi\in S$ and it is equal one near the origin, then $||\phi(\cdot/R)T_1(\phi(\cdot/R)f)(x)||_p=R^{d/p}||(\phi T_R\phi)f_R||_p\le CR^{d/p}||f_R||_p=C||f||_p$, where $f_R=f(R\cdot)$. Note that I used the uniform boundedness for $\phi T_R\phi$ and the fact the $\phi(\cdot/R)f\to f$ in $L^p$

The way you should understand $T_1$ is as the extension of the operator $T_1f=(\hat{f}\chi_{B(0,1)})^{\vee}$ for $f\in S$, whenever you have $L^p$-boundedness. Since $\hat{f}\chi_{B(0,1)}\in L^1$ for $f\in S$, there is no problem with the definition. A convolution could belong to $L^p$, despite the "bad" behaviour of one of the functions, for example $f*\delta = f$.

I hope I have done everything well, I am not sure.

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  • $\begingroup$ Thank you! This clears up my confusion. $\endgroup$
    – Siddhant
    Mar 29, 2015 at 5:46

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