Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
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 7 '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 7 '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 7 '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 7 '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 7 '11 at 14:34
up vote 3 down vote accepted

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.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.