Properties of family of translation operators 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.
 A: 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.
