Translation-Invariant Operators on Lebesgue Spaces

[Euge]

Show that if there exists a nonzero, translation-invariant bounded linear operator ##T : L^p(\mathbb{R}^d) \to L^q(\mathbb{R}^d)## where ##1 \le p, q < \infty##, then necessarily ##q \ge p##.

 

[Office_Shredder]

Does translation invariant mean that ##T(f(x+c))=T(f(x))## for any constant c?

 

[Euge]

Does translation invariant mean that ##T(f(x+c))=T(f(x))## for any constant c?

If ##y\in \mathbb{R}^d## and ##(\tau_y f)(x) := f(x + y)## for all ##x\in \mathbb{R}^d##, we require ##T\circ \tau_y = \tau_y\circ T## for all ##y\in \mathbb{R}^d##.

 

[Office_Shredder]

If ##y\in \mathbb{R}^d## and ##(\tau_y f)(x) := f(x + y)## for all ##x\in \mathbb{R}^d##, we require ##T\circ \tau_y = \tau_y\circ T## for all ##y\in \mathbb{R}^d##.

Thanks. I tried to Google translation invariant operator but the first hit was a proof of this result, so I stopped looking.

 

[Euge]

Spoiler

If ##f : \mathbb{R}^d \to \mathbb{R}## is a smooth and compactly supported, the supports of ##f## and ##\tau_y f## are disjoint for all sufficiently large ##|y|##. Hence ##\|f + \tau_yf\|_p = (\|f\|_p^p + \|\tau_yf\|_p^p)^{1/p} = 2^{1/p}\|f\|_p## for ##|y| \gg 1##. By translational invariance of ##T##, a similar argument shows ##\|T(f + \tau_yf)\|_q = 2^{1/q} \|Tf\|_q## for ##|y| \gg 1##. Boundedness of ##T## forces ##2^{1/q} \|Tf\|_q \le \|T\| 2^{1/p} \|f\|_p##. Replacing ##f## with ##f + \tau_yf## and repeating the analysis produces the inequality ##2^{2/q} \|Tf\|_q \le \|T\| 2^{2/p} \|f\|_p##. Repeating the process, we find ##2^{N/q} \|Tf\|_q \le \|T\| 2^{N/p} \|f\|_p## for all positive integers ##N##. By density of smooth compactly supported functions in ##L^p## the same inequalities hold true for all ##f\in L^p##.

Since ##T## is nonzero, there is a nonzero ##f\in L^p## such that ##\|Tf\|_q > 0##. The inequality $$\|T\| \ge 2^{N(1/p - 1/q)} \frac{\|Tf\|_q}{\|f\|_p}\quad (N = 1,2,3,\ldots)$$ forces ##1/p - 1/q \le 0## (otherwise ##\|T\| = \infty##). In other words, ##q \ge p##.