Translations in $L^p(\mathbb{R}^n)$

Question

Let $f,g\in L^p(\mathbb{R}^n)$, $1\leq p< +\infty$. Define (for a.e. $x\in\mathbb{R}^n$) $g_h(x)=g(x-h)$. Show that $$ \lim_{h\to\infty} \lVert f-g_h \rVert_p=(\lVert f\rVert_p^p+\lVert g\rVert_p^p)^{1/p}. $$

Any suggestions?

Answer 1

It works for functions of compact support; if the supports of $f,g$ are contained in $B_R(0)$, the support of $g_h$ is contained in $B_R(h)$. These are disjoint for $\| h \| \geq 2R$, so for $\| h \| \geq 2R$, $$\begin{align} \int |f(x)+g_h(x)|^p\, dx &= \int_{B_R(0)}|f(x)|^p \, dx + \int_{B_R(h)}|g(x-h)|^p\, dx \\ &= \int_{B_R(0)} |f(x)|^p\, dx + \int_{B_R(0)} |g(x)|^p\, dx \\ &= \|f\|_p^p +\|g\|_p^p,\end{align}$$ now take $p$'th roots.

To extend to the general case; it follows from the density of smooth functions of compact support in $L^p$ (though some care must be taken to "swap the limits"); if you approximate $f$ and $g$ in $L^p$ by say $f_\epsilon$ and $g_\epsilon$ then $f_\epsilon - (g_\epsilon)_h$ will approximate $f-g_h$ in $L^p$ by the triangle inequality and translation invariance; $$\begin{align}\|(f_\epsilon - (g_\epsilon)_h)-(f-g_h)\|_p &\leq \| f_\epsilon - f \|_p + \|(g_\epsilon)_h - g_h\|_p \\ &= \|f_\epsilon - f\|_p + \|g_\epsilon - g\|_p.\end{align}$$

Being precise; let $\epsilon >0$. Because $(x,y) \mapsto (x^p + y^p)^{1/p}$ is continuous at $(\|f\|_p,\|g\|_p)$, there exists $\delta > 0$ such that if $(x,y)$ is within $\delta$ of $(\|f\|_p,\|g\|_p)$, then $(x^p+y^p)^{1/p}$ is within $\epsilon$ of $(\|x\|_p^p + \|y\|_p^p)^{1/p}$. Now take $f_\epsilon$, $g_\epsilon$ of compact support with $$\| f- f_\epsilon\|_p, \|g-g_\epsilon\|_p < \min(\delta/2,\epsilon).$$ Now the reverse triangle inequality $| \| f \|_p - \|f_\epsilon \||_p \leq \| f- f_\epsilon \|_p$ (and the same for $g$) implies $(\|f_\epsilon\|_p,\|g_\epsilon\|_p)$ is within $\delta$ of $(\|f\|_p,\|g\|_p)$, hence $(\|f_\epsilon\|_p^p+\|g_\epsilon\|^p_p)^{1/p}$ is within $\epsilon$ of $(\|f\|_p^p + \|g\|_p^p)^{1/p}$. Then we get $$\| f_\epsilon - (g_h)_\epsilon \|_p - 2 \epsilon \leq \| f - g_h \|_p \leq \|f_\epsilon - (g_\epsilon)_h\|_p + 2\epsilon,$$ which gives $$(\|f_\epsilon\|_p^p + \|g_\epsilon\|_p^p)^{1/p} -2\epsilon\leq \liminf_{\|h \| \rightarrow \infty}\|f - g_h\|_p \leq \limsup_{\|h \| \rightarrow \infty} \|f - g_h \|_p \leq (\|f_\epsilon\|_p^p + \|g_\epsilon\|_p^p)^{1/p} + 2\epsilon.$$ Then this gives $$(\|f\|_p^p + \|g\|_p^p)^{1/p} - 3\epsilon \leq \liminf_{\|h \| \rightarrow \infty}\|f - g_h\|_p \leq \limsup_{\|h \| \rightarrow \infty} \|f - g_h \|_p \leq (\|f\|_p^p + \|g\|_p^p)^{1/p} + 3\epsilon,$$ and as $\epsilon$ is arbritrary we are done.

Answer 2

The main idea is : \begin{eqnarray*} ||f+ g||_p^p= ||f||_p^p + ||g||_p^p \end{eqnarray*} if the supports of $f$ and $g$ are disjoint.

Approximate $f$ and $g$ by $f \cdot \chi_{B_R}$ and $g \cdot \chi_{B_R}$. Then use the fact that for $a$ large enough $f \cdot \chi_{B_R}$ and $T_a (g \cdot \chi_{B_R})$ have disjoint supports and also that $T_a$ is an isometry.