Translation-Invariant Operators on Lebesgue Spaces

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In summary, translation-invariant operators on Lebesgue spaces are linear operators that preserve the structure of functions under translations. Unlike other operators, they only change the translation component of the function or data being operated on. Common examples include convolution, Fourier, and wavelet transform operators, which are useful in fields such as signal and image analysis. They allow for the detection of consistent patterns and features under translations, which can be applied in practical applications such as image and speech recognition. However, they may not be suitable for all types of functions or data, particularly those without a well-defined translation structure.
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Euge
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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##.
 
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Does translation invariant mean that ##T(f(x+c))=T(f(x))## for any constant c?
 
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Office_Shredder said:
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##.
 
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Euge said:
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.
 
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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##.
 

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