POTW Translation-Invariant Operators on Lebesgue Spaces

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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 $q \ge p$.

 

[Office_Shredder]

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

 

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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.

 

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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, $q \ge p$.