Show strong but not norm convergence on $L^p$

Question

The task:

let $1 \le p < \infty$ and $A_k : L^p (\Bbb R) \to L^p (\Bbb R)$ such that $(A_k u) (x) = u(x+\frac 1 k)$. Show that $\| A_k u - u\|_p \to 0$ as $k \to \infty$ (for all $u \in L^p (\Bbb R)$), but that $\| A_k - I \| \not\to 0$ as $k \to \infty$ (in other words, $A_k$ converges strongly to $I$, but not in norm).

I think I managed to do the first part rather weakly by mimicking this one here but I am interested in how you can show rigorously that $|| A_{k} - I || \not\to 0$.

My draft proposal to whole task

Recall that \begin{equation*} A_{k} u(x) := u \left( x + \frac{1}{k} \right), \end{equation*} \begin{equation*} || u ||_{p} := \left( \sum\limits_{k \in \mathbb{Z} } | u_{k} |^{p} \right)^{1/p}, \, \forall u = \{ u_{k} \}_{k \in \mathbb{Z} } \in l_{p}, \, 1 \leq p < \infty. \end{equation*}

Let an element be $u \in l^{p}(\mathbb{R})$, and a sequence of elements $(v^{k})_{k=1}^{\infty} \subset l^{p}(\mathbb{R})$. The series \begin{equation*} ||u||_{p}^{p} = \sum\limits_{k \in \mathbb{Z}} |u_{k}|^{p} < \infty, \end{equation*} is (absolutely) convergent which implies that its tail tends to zero: \begin{equation*} \sum_{ |k| > n } |u_{k}^{p}| \to 0 \, \text{ when } \, n \to \infty. \end{equation*} Take for approximating sequence $\{ A^{k} \}_{k=1}^{\infty} := \{ v^{k} \}_{k=1}^{\infty} \subset l^{p}(\mathbb{R}):$ \begin{equation*} v^{k} = (0,0, ..., 0, u_{-k}, u_{-k+1}, ..., u_{-1}, u_{0}, u_{1}, ..., u_{k-1}, u_{k}, 0, 0, ...) \in l^{p}( \mathbb{R} ) \end{equation*} and see that the sequence converges to $u$ for all $u \in L^{p}(\mathbb{R})$: \begin{equation*} || v^{k} - u ||_{p} = \sum\limits_{ |n| > k } | u_{n} |^{p} \to 0 \, \text{ when } \, k \to \infty, \end{equation*} Thus $||A_{k} u - u ||_{p} \to 0$ as $k \to \infty. \, \square$

Consider the tail of the mapping when $n = k$: \begin{align*} || A_{k} - I ||_{p} &= || u \left( x + \frac{1}{k} \right) - I || \\ &= || \frac{u \left( x + 1/k \right)^{2} - u(x + 1/k)}{ u(x + 1/k) } || \not\to 0 \, \text{ when } \, k \to \infty, \end{align*} because $u(x + 1/k) \neq 0.$ $A_{k}$ converges strongly to $I$, but not in norm. $\square$

Comments:

• I think $A_{k} u(x)$ means the $k$-th term. Does it?
• Taking a sequence in the domain but not in the subset of the domain worries me

Rewritten Martini's answer

Recall that \begin{equation*} A_{k} u(x) := u \left( x + \frac{1}{k} \right), \end{equation*} To prove strong convergence, the space of continuous maps with the compact support, $C_c(\mathbb{R})$, is dense in $L^p(\mathbb{R})$. For the element $u \in C_c(\mathbb{R})$, $| u - u_{k} |$ is bounded above by an integrable function (reverse Fatou lemma): $\| u - A_{k} u \|$ is bounded by $2\|u\|_{\infty} \cdot 1_{\text{supp} \, u+(-1,1)},$ where $1_{A}$ denotes the characteristic function of a set $A$, and \begin{equation*} \lim\limits_{n \to \infty} \sup \int_{\mathbb{R}} | u - u_{k}| \leq \int_{S} \lim\limits_{n \to \infty} \sup | u - u_{n} | d\mu = 0, \end{equation*} which implies that the limit exists and vanishes: \begin{equation*} \lim\limits_{n \to \infty} \int_{\mathbb{R}} | u - u_{n} | d\mu = 0. \end{equation*} which implies that by the dominated convergence theorem: \begin{equation*} || A_k u - u ||_p^p = \int_{\mathbb{R}} | u\left(x + \frac{1}{k} \right) - u(x) |^p \, dx \to 0, \qquad k \to \infty \end{equation*} where the continuity of $u$ gives the pointwise convergence of the integrand. ... (rest in Martini's answer)

Comments:

• the first part of the proof have some things about which I am not sure
• the second part is ok by the proof of contradiction

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How can you show that $A_{k}$ does not converge strongly to $I$ in the norm?

Answer 1

Hint. $A_k u$ is by definition the map $x \mapsto u(x + 1/k)$ (where $u \colon \mathbf R \to \mathbf R$ is a given measurable, $p$-integrable map). You only have a sequence of maps $A_k \colon L^p(\def\R{\mathbf R}\R) \to L^p(\R)$ here, neither the domain nor its subsets consist of sequences, $L^p(\R)$ is the space of $p$-integrable maps $\R \to \R$. Do not confuse it with $\ell^p$, which consists of sequences.

To prove strong convergence, recall that $C_c(\R)$, the space of continuous maps with compact support, is dense in $L^p(\R)$. For $u \in C_c(\R)$, use the dominated convergence theorem to show that $$ \|A_k u - u\|_p^p = \int_{\R} \def\abs#1{\left|#1\right|}\abs{u\left(x + \frac 1k \right) - u(x)}^p\, dx \to 0, \qquad k\to \infty $$ (the continuity of $u$ gives pointwise convergence of the integrand). For arbitrary $u \in L^p(\R)$, given $\epsilon > 0$, choose $v \in C_c(\R)$ with $\|u - v\|_p < \epsilon$. As, by translation invariance of the integral $\|A_k u - A_k v\|_p = \|u - v\|_p$, we have $$ \|u - A_k u\|_p \le 2\epsilon + \|v- A_k v\|_p $$ and the latter converges to zero.

To disprove norm convergence, note that for any $k \in \mathbf N$, there is a function $u \in L^p(\R)$ such that $u$ and $A_k u$ have disjoint support and $\|u\|_p = 1$. For example, let $u = k^{1/p}\chi_{[0,\frac 1k)}$. We have $$ \|A_k - I\|_p \ge \|A_k u - u\|_p = 2\|u\|_p = 2 $$

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Addendum: If you want to avoid using the dominated convergence theorem, you can argue as follows: Let $u \in C_c(\R)$ and $\epsilon > 0$ be given. Choose $M > 0$ such that $u(x) = 0$ for $x \not\in [-M,M]$. As $u$ is uniformly continuous (as it has compact support), choose $\delta > 0$ such that $|x- y| < \delta$ implies $|u(x) - u(y)| < {\epsilon}/{(2M+2)^{1/p}}$. For $k > \frac 1\delta$, we have $$ \|{A_k u - u}\|^p_p = \int_{\R} \abs{u\left(x+\frac 1k\right)-u(x)}^p\, dx \le \int_{-M-1}^{M+1} \frac{\epsilon^p}{2M+2}\, dx = \epsilon^p $$ that is $\|A_k u - u\|_p \le \epsilon$.