Let $f\in L^p (R)$ for $1\leq p<\infty$ and let $f_t(x)=f(x-t)$. Let $f\in L^p (R)$ for $1\leq p<\infty$ and let $f_t(x)=f(x-t)$.


*

*Prove each $f_t\in L^p(R)$.

*Prove $\lvert \lvert f_t-f\lvert\lvert_p\rightarrow0$ as $t\rightarrow0$. What happens as $t\rightarrow \infty$?


So the first question I’m letting $y=x-t$, since $f\in L^p (R)$ implies $\int \lvert f (x)\lvert dx <\infty $,$\int \lvert f(y) \lvert dy <\infty$.Then by substitution we have, $\int \lvert f (x-t)\lvert dx <\infty $, and we are done.
But I’m not sure about the second question. Any idea on how to bound that term? Thanks.
 A: 
Let $\epsilon > 0$ and $\varphi \in C_{c}^{\infty}(\mathbb{R})$ be such that $\|f-\varphi \|_{p} < \epsilon/3$. Because $\varphi$ has compact support and continous functions on compact sets are uniformly continuous, $\varphi$ is uniformly continuous. It follows that there is a $\delta > 0$ such that $\|\varphi - \varphi_{t}\|_{L^p} < \epsilon/3$ for all $t < \delta$. We find
$$\|f-f_{t}\|_{p} \leq \|f-\varphi\|_{p} + \|\varphi - \varphi_{t}\|_{p} + \|\varphi_{t}-f_{t}\|_{p} \leq \epsilon,$$
for $t < \delta$, so $\|f_{t}-f\| \to 0$ as $t \to 0$.

Alternatively, you could show that the space of all functions $f \in L^{p}(\mathbb{R})$ such that $\|f-f_{t}\|\to 0$ as $t\to\infty$ is a closed vector space that contains the indicator functions of compact intervals.

Regarding what happens as $t \to \infty$, consider $f(x) := 1_{[0,1]}(x)$. Then $f_{t}(x) = 1_{[t,t+1]}(x)$, which is not Cauchy in $L^p(\mathbb{R})$, so the limit of $(f_{t})_{t\geq0}$ as $t\to\infty$ does not exist. Hence, we can not in general hope to find the limit of the translations as $t$ goes to infinity (in $L^{p}(\mathbb{R}))$.
