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For a family of linear operators $U_y : L^p(\mathbb{R}) \rightarrow L^p(\mathbb{R}) $ defined by $U_y f(x) = f(x + y)$ and $p \in [1, \infty]$ I'm asked to prove the following:

a) For any fixed $y \in \mathbb{R}$ the operator $U_y : L^p \rightarrow L^p$ is a bounded linear operator on $L^p$ for any $p \in [1, \infty]$. What is its norm?

My answer:

claim 1: $U_y$ is bounded

proof 1: $$ \int_R |f(x + y ) |^p dx = \int_R |f(x ) |^p dx \implies \| U_y f \|_p = \| f \|_p < \infty $$

claim 2: $\| U_y \|_{op} = 1$

proof 2: $$ \| U_y \|_{op} = \sup_{f \in L^p, \| f \| \leq 1} \| U_y f \|_{op} = \sup_{f \in L^p, \| f \| \leq 1} \| f \|_{op} = 1$$

Is this right? And b):

b) Fix $f \in L^p$ and consider the map $\mathbb{R} \rightarrow L^p$, $y \mapsto U_y (f)$. For which $p$ is this map continuous?

I thought that if I have $B_\varepsilon ( f(x + y)) \subset L^p$, then for $f(x + x_0) \in B_\varepsilon ( f(x + y))$:

$$ \| f(x + x_0) - f(x + y) \|_p < \varepsilon $$

So using $$\int_R |f(x + y ) |^p dx = \int_R |f(x ) |^p dx $$ again I would get

$$ \| f(x + x_0) - f(x + y) \|_p = \| f(x) - f(x ) \|_p = 0 < \epsilon$$ so I could pick any $\delta$ and the map, let's call it $F$, would be continuous for all $p$. What do you think of this?

Many thanks for your help.

Rudy the Reindeer
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    proof 1 already shows that the translation operator is in fact an isometry for all $p$. Concerning b) I don't understand what you're doing here at all. In fact, you have continuity for all $p \lt \infty$. It fails for $p = \infty$ (take $f$ to be a characteristic function of an interval to see why). See this answer for some inspiration (the argument works for all $p \lt \infty$, as the continuous functions of compact support are dense in $L^p$ provided $p \lt \infty$). – t.b. Nov 07 '11 at 13:09
  • @t.b.: Thank you. In b) I am trying to use the $\varepsilon \delta$ definition of continuity. Now I'm reading Jonas' answer: do I really need $f$ to have compact support? I think continuity is enough. – Rudy the Reindeer Nov 07 '11 at 13:22
  • Jonas uses uniform continuity which you don't have for continuous functions. ad b) Well, I see what you're trying to do, but, as I said, I can't make much sense out of it. Why do you find $f(x+x_0)$ in that ball in the first place? What happens after "again I would get"? – t.b. Nov 07 '11 at 13:33
  • @t.b.: I cannot make any sense of it either. But regarding J's answer: I was referring to "Note that ... pointwise". I still think that it only needs continuity. – Rudy the Reindeer Nov 07 '11 at 14:10
  • Yes, sure, you have this pointwise convergence from continuity only. However, as you should know, this is not enough to ensure the convergence of the integrals you compute when considering $|f - U_y f|_p$: you need to apply dominated convergence or something like that, but it's not quite obvious how to do that. Uniform continuity of $f$ makes this easier. Don't miss Julián's comment to the answer! – t.b. Nov 07 '11 at 14:34
  • By the way, you may also want to look at this answer. – t.b. Nov 07 '11 at 14:45
  • @t.b.: Thank you! I think it will help me too but I'm still reading the first link you gave me. : ( I should've finished 4 questions today and I haven't even finished one. – Rudy the Reindeer Nov 07 '11 at 15:00
  • @t.b.: I finished reading and almost finished understanding. Why does he need the last step? Can I not make $| g - f_n | $ arbitrarily small already because of $C^c (X) $ dense in $L^1$? Or isn't $f_n \in L^1$? Is it o.k. to ask you this in a comment or should I make this a question? – Rudy the Reindeer Nov 07 '11 at 15:29
  • No problem, but I don't understand what you're asking. The very last sentence is a general remark about attacking such questions. The last step in the answer is an $\varepsilon/3$-argument. Make $|f - g| \lt \varepsilon/3$, then $|f_n - g_n| \lt \varepsilon /3$ by translation invariance, while $|g-g_n| \to 0$ by the first part of the argument. See also my answer I linked to above. – t.b. Nov 07 '11 at 15:35
  • @t.b.: I just made a terrible mistake. Again. I wish I would be making less basic mistakes over and over again. Thank you for being so patient! Time to read the second link. – Rudy the Reindeer Nov 07 '11 at 15:56

1 Answers1

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Addressing question b), fix $p \in [1,\infty)$. For $p = \infty$, as pointed out in the comments by t.b., take a suitable characteristic function.

Claim 1. Let $f \in C_c(\mathbb{R})$. Then the map $$ \mathbb{R} \rightarrow C_c(\mathbb{R}), y \mapsto U_y f $$ is uniformly continuous with respect to the norm $\|{\cdot}\|_{C_c} = \max_x |f(x)|$.

Proof. Since $f$ is continuous and compactly supported we have that $f$ is uniformly continuous. Hence for all $\varepsilon > 0$ we can choose $\delta$ such that for all $x, y, y' \in \mathbb{R}$ we have

$$|f(x+y)-f(x+y')| < \varepsilon$$

if $\|x+y-(x+y')\| = \|y-y'\| < \delta$. This implies that $\|U_yf - U_{y'}f\|_{C_c} < \varepsilon$ for the $\delta$ above.

Claim 2. Let $f \in L^p(\mathbb{R})$. Then the map $$ \mathbb{R} \rightarrow L^p(\mathbb{R}), y \mapsto U_y f $$ is uniformly continuous.

Proof. Let $\varepsilon > 0$ and let $y' \in \mathbb{R}$. Since $C_c(\mathbb{R})$ is dense in $L^p(\mathbb{R})$ for the given values of $p$ we can choose $g \in C_c(\mathbb{R})$ such that

$$ \|f-g\|_p < \varepsilon / 3 $$

Now, using linearity from a), write $$ U_yf - U_{y'} f = U_y g - U_{y'}g - U_y(g-f) + U_{y'}(g-f) $$

Put $S := \mathrm{supp}(g)$. We have that

$$ \mathrm{supp} (U_yg - U_{y'}g) \subset \{x+y; x\in S\} \cup \{x+y'; x \in S\} $$

It follows that

$$\lambda (\mathrm{supp} (U_yg - U_{y'}g)) \leq 2 \lambda(S),$$

where $\lambda$ denotes the Lebesgue measure on $\mathbb{R}$ and using its translation invariance. With this we get

$$ \|U_yg-U_{y'}g\|_p \leq \max |U_yg - U_{y'}g| \cdot 2\lambda(S) = \|U_yg-U_{y'}g\|_{C_c} \cdot 2 \lambda (S). $$

By claim 1 choose a $\delta > 0$ such that $$ \|U_yg-U_{y'}g\|_{C_c} < \varepsilon / (3\cdot 2\lambda(S)) $$

For this $\delta$ we have

$$ \|U_y f -U_{y'} f\|_p < 2\varepsilon / 3 + \varepsilon / 3 = \varepsilon $$

and thus the desired claim.

NB: This is my first answer, so I am especially grateful for any inputs.

caligula
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