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For every $a \in \mathbb{R}^d$, let $T_a(f)(x) = f(x-a), \forall f \in L^p(\mathbb{R}^d), \forall x \in \mathbb{R}^d$. Prove that $T_a$ is a linear isometry of space $L^p$ on itself. Find $\lim_{a\to 0} T_a$ if it exists (uniform, strong, and weak). Does the result depend on $1 \leq p \leq \infty$?

I have proved the first part, $$ ||T_a(f)|| = \left(\int_{R^d}|f(x-a)|^p dm_d(x)\right)^{1/p} = \left(\int_{R^d}|f(x)|^p dm_d(x)\right)^{1/p} = ||f||, $$ because of the invariance of the Lebesgue measure to translation and central symmetry. But I don't know what to do with the rest. Thanks in advance!

qianqian07
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Maria
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  • Strong convergence is a standard result. You can find it in Rudin's RCA. Uniform convergence fails: look at characteristic functions of intervals. – geetha290krm Mar 19 '23 at 12:31
  • I do not see here what is $lim_{a\to 0} T_a$, I see only that $\lim_{a\to 0} T_a(f) = f$. – Maria Mar 19 '23 at 12:51

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  1. Strong convergence means that for each $f$, $T_af \rightarrow T_0f$ in $L^p$. This is obvious since $f(x-a) \rightarrow f(x) \in L^p$.

  2. Strong convergence implies weak convergence. So $T_a$ also converges weakly to $T_0$.

  3. Norm convergence fails. Consider $||T_a - T_0|| = sup_{||f||=1} ||T_a(f) - T_0(f)||$. However, we can choose $f$ to be a function that is supported in $B_a/2(0)$. Then $f(x)$ and $f(x-a)$ has disjoint support and $||T_a(f) - T_0(f)|| > ||f||$.

Alvis Zhalovsky
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