$L^{p}$ convergence equivalent condition

Question

I have to show that for $p\in[0,\infty)$, $f_{n},f\in L^{p}(\mathbb{R})$, (i) $f_n\rightarrow f$ in $L^{p}([-N,N])$ for all $N\in \mathbb{N}$ and (ii) $\lvert\lvert f_{n}\rvert\rvert \rightarrow \lvert\lvert f\rvert\rvert$ implies $f_{n}\rightarrow f$ in $L^{p}(\mathbb{R})$.

My approach was to first deconstruct $\lvert\lvert f_{n}-f\rvert\rvert_{p}^{p}$ as $$ \lvert\lvert f_{n}-f\rvert\rvert_{p}^{p}=\int_{-\infty}^{-N}\lvert f_{n}-f\rvert^{p}+\int_{-N}^{N}\lvert f_{n}-f\rvert^{p}+\int_{N}^{\infty}\lvert f_{n}-f\rvert^{p}. $$ (The integral is taken w.r.t. the Lebesgue measure). By (i) we know that the above decomposition is valid for all $N\in \mathbb{N}$ and as we take the limit $n\rightarrow \infty$, the middle term vanishes. The problem is how to get an upper bound $\epsilon_{n}$ for the left and right integral s.t. $\epsilon_{n}\rightarrow 0$ as $n\rightarrow \infty$. I don't see how we could use (ii) for that. I tried to show that for all $n$ sufficiently large we get an $N^{*}$ such that the left and right integral are bounded and that this bound approaches $0$ as $n \rightarrow \infty$, but didn't succeed. Could anyone help?

Answer 1

We can show it using the following facts.

1. If $\left(g_n\right)_{n\geqslant 1}$ is a sequence such that $g_n\to g$ almost everywhere and $\int \lvert g_n\rvert^p\to\int \lvert g\rvert^p$, then $\int \lvert g_n-g\rvert^p\to 0$.

2. If $(f_n)_{n\geqslant 1}$ satisfies the conditions of the opening post, then there exists a subsequence $(g_n)=(f_{\varphi(n))})$ where $\varphi\colon\mathbb N\to\mathbb N$ is increasing, such that $\int \lvert g_n-f\rvert^p\to 0$.

In order to show fact one, we can use Fatou's lemma to the sequence $(h_n)$ given by $$ h_n= \begin{cases} \lvert f\rvert^p+\lvert f_n\rvert^p-\lvert f_n-f\rvert^p&\mbox{ if }0<p\leqslant 1,\\ 2^{p-1}\lvert f\rvert^p+2^{p-1}\lvert f_n\rvert^p-\lvert f_n-f\rvert^p&\mbox{ if } p\geqslant 1. \end{cases} $$

For fact 2., we use the fact that convergence in $L^p$ implies the almost sure convergence of a subsequence combined with a diagonal extraction argument.

To conclude, fact 2. applied to a subsequence of $(f_n)$, say $(f_{\psi(n))})$ show that there exists a further subsequence $(f_{\varphi\circ \psi(n)})$ such that $\int\left\lvert f_{\varphi\circ \psi(n)}-f\right\rvert^p\to 0$.

Answer 2

Let $\epsilon > 0$, and choose $N$ so large that $\|f\|_{L^p(\mathbb R\setminus [-N,N])} < \epsilon$. (Why does such an $N$ exist?)

Note that (i) $\|f-f_n\|_{L^p([-N,N])}\to 0$ implies $\|f_n\|_{L^p([-N,N])}\to \|f\|_{L^p([-N,N])}$. Then (ii) $\|f_n\|_p\to \|f\|_p$ and $\|f_n\|_{L^p([-N,N])}\to \|f\|_{L^p([-N,N])}$ together imply $$ \|f_n\|_{L^p(\mathbb R\setminus[-N,N])}^p = \|f_n\|_p^p - \|f_n\|_{L^p([-N,N])}^p\to \|f\|_{L^p(\mathbb R\setminus [-N,N])}^p. $$ So $$ \|f-f_n\|_p^p = \|f-f_n\|_{L^p([-N,N])}^p + \|f-f_n\|_{L^p(\mathbb R\setminus [-N,N])}^p, $$ and sending $n\to\infty$ the limit superior of the right-hand side is bounded by $(2\|f\|_{L^p(\mathbb R\setminus [-N,N])})^p < (2\epsilon)^p$.

Answer 3

Choose $M>0$ such that $\int|f_{n}|^{p}\leq M$ for all $n$. (This is possible becuase $\int|f_{n}|^{p}\rightarrow\int|f|^{p}<\infty$). Note that we also have $\int|f|^{p}\leq M$. Let $\varepsilon>0$ be arbitrary. Since $f\in L^{p}$, we may choose $N\in\mathbb{N}$ such that $\int_{[-N,N]^{c}}|f|^{p}<\varepsilon$ (This follows from Dominated Convergence Theorem by observing that $1_{[-N,N]^{c}}(x)|f(x)|^{p}\rightarrow0$ as $N\rightarrow\infty$ and that it is dominated by the integrable function $|f|^{p}$). Since $\int_{[-N,N]}|f_{n}-f|^{p}\rightarrow0$ as $n\rightarrow\infty$, we may choose $N_{1}\in\mathbb{N}$ such that $\int_{[-N,N]}|f_{n}-f|^{p}<\varepsilon^{p}$ whenever $n\geq N_{1}$. Choose $N_{2}\in\mathbb{N}$ such that $\left|\int|f_{n}|^{p}-\int|f|^{p}\right|<\varepsilon$ whenever $n\geq N_{2}$.

Let $n\geq\max(N_{1},N_{2})$. We go to prove that $\int_{[-N,N]^{c}}|f_{n}|^{p}$ is small. We have estimation: \begin{eqnarray*} & & \int_{[-N,N]^{c}}|f_{n}|^{p}+\int_{[-N,N]}|f_{n}|^{p}\\ & = & \int|f_{n}|^{p}\\ & \leq & \int|f|^{p}+\varepsilon\\ & = & \int_{[-N,N]^{c}}|f|^{p}+\int_{[-N,N]}|f|^{p}+\varepsilon.\\ & \leq & 2\varepsilon+\int_{[-N,N]}|f|^{p}. \end{eqnarray*} Therefore, $\int_{[-N,N]^{c}}|f_{n}|^{p}\leq2\varepsilon+\int_{[-N,N]}|f|^{p}-\int_{[-N,N]}|f_{n}|^{p}$. To simplify notation, we denote $||g||=\left\{ \int_{[-N,N]}|g|^{p}\right\} ^{\frac{1}{p}}$ for any measurable function $g:\mathbb{R}\rightarrow\mathbb{R}$. We have estimation: \begin{eqnarray*} & & ||f||\\ & = & ||(f-f_{n})+f_{n}||\\ & \leq & ||f-f_{n}||+||f_{n}||\\ & \leq & \varepsilon+||f_{n}||. \end{eqnarray*} Let $\theta:[0,\infty)\rightarrow\mathbb{R}$ be defined by $\theta(x)=x^{p}$. By mean-value theorem, we have that \begin{eqnarray*} \theta(||f||)-\theta(||f_{n}||) & = & \theta'(\xi)\left\{ ||f||-||f_{n}||\right\} \\ & \leq & \varepsilon\theta'(\xi) \end{eqnarray*} where $\xi$ is a number between $||f||$ and $||f_{n}||$, and hence $\xi\leq M^{\frac{1}{p}}$. It follows that $\theta'(\xi)=p\xi^{p-1}\leq p\cdot M^{\frac{p-1}{p}}.$ Therefore, \begin{eqnarray*} \int_{[-N,N]^{c}}|f_{n}|^{p} & \leq & 2\varepsilon+||f||^{p}-||f_{n}||^{p}\\ & \leq & 2\varepsilon+\varepsilon pM^{\frac{p-1}{p}}. \end{eqnarray*} Furthermore, we have that \begin{eqnarray*} \left\{ \int_{[-N,N]^{c}}|f_{n}-f|^{p}\right\} ^{\frac{1}{p}} & \leq & \left\{ \int_{[-N,N]^{c}}|f_{n}|^{p}\right\} ^{\frac{1}{p}}+\left\{ \int_{[-N,N]^{c}}|f|^{p}\right\} ^{\frac{1}{p}}\\ & \leq & \left\{ 2\varepsilon+\varepsilon pM^{\frac{p-1}{p}}\right\} ^{\frac{1}{p}}+\varepsilon^{\frac{1}{p}}. \end{eqnarray*}

Finally, for $n\geq\max(N_{1},N_{2})$, we have estimation \begin{eqnarray*} & & \int|f_{n}-f|^{p}\\ & = & \int_{[-N,N]}|f_{n}-f|^{p}+\int_{[-N,N]^{c}}|f_{n}-f|^{p}\\ & \leq & \varepsilon^{p}+\left[\left\{ 2\varepsilon+\varepsilon pM^{\frac{p-1}{p}}\right\} ^{\frac{1}{p}}+\varepsilon^{\frac{1}{p}}\right]^{p}. \end{eqnarray*} It follows that $\int|f_{n}-f|^{p}\rightarrow0$ as $n\rightarrow\infty$.