How can I show that this piecewise linear map $T$ is continuous? Let $V=L^1([0,2])$ be space of integrable functions, equipped with the norm:
$$\lVert f(x)\rVert=\int_0^2\vert f(x) \vert dx$$
Consider the linear mapping:
$$T: V \to V, [Tf](x):=\begin{cases} f(x+1) & \text{for} &x\in(0,1] \\f(x-1) & \text{for} & \text{elsewhere} \end{cases}$$
Question: How can I show that this is continuous? For a piecewise function I would just check if the values of the two functions at the point $x=1$ is the same. However, I don't have explicit functions here. What to I need to do to show $T$ is continuous or not?
Sample solution from an old exam (I am just not sure whats being done here):


a) $$\lVert T\rVert=\sup_{\substack{f\neq0\\ f\in V}}\frac{\lVert Tf\rVert}{\lVert f\rVert}=\sup\frac{\int_0^1\left\lvert f\left(x+\frac{1}{2}\right)\right\rvert~\mathrm{d}x+\int_1^2\left\lvert f\left(x\right)\right\rvert~\mathrm{d}x}{\int_0^2\left\lvert f\left(x\right)\right\rvert~\mathrm{d}x}\leq2\cdot\frac{\int_0^2\left\lvert f\left(x\right)\right\rvert~\mathrm{d}x}{\int_0^2\left\lvert f\left(x\right)\right\rvert~\mathrm{d}x}=2<\infty$$ $\implies$ continuous
b) $\sup$: find function s.t. $\lVert T\rVert$ is maximal: $$f_m(x)=\begin{cases}\lambda,&1\leq x\leq \frac{3}{2},\\ 0&\text{else},\end{cases}$$ $\lambda\in\mathbb{R}$, $\lambda\neq0$. $$\lVert T\rVert=\frac{\lVert Tf_m\rVert}{\lVert f_m\rVert}=\frac{\int_0^1\left\lvert f_m\left(x+\frac{1}{2}\right)\right\rvert~\mathrm{d}x+\int_1^2\left\lvert f_m\left(x\right)\right\rvert~\mathrm{d}x}{\int_0^2\left\lvert f_m\left(x\right)\right\rvert~\mathrm{d}x}=\frac{\lvert\lambda\rvert\cdot0.5+\lvert\lambda\rvert\cdot 0.5}{\lvert\lambda\rvert\cdot0.5}=2$$

 A: A linear map $T:E\to F,$ where $E,F$ are normed vector spaces, is continuous iff
$$\exists C\in\Bbb R\quad\forall f\in E\quad\|T(f)\|_F\le C\|f\|_E$$
(it is a theorem, not a definition).
Here, $E=F=L^1([0,2])$ and $\|f\|=\int_0^2|f(x)|\,dx.$
$$\|T(f)\|=\int_0^2|(Tf)(x)|\,dx=\int_0^1|f(x+1)|\,dx+\int_1^2|f(x-1)|\,dx$$
$$=\int_1^2|f(u)|\,du+\int_0^1|f(v)|\,dv=\|f\|$$
(i.e. $T$ is an isometry), so you may take $C=1.$
A: You could just go down to the definition of continuity:

$T$ is continuous if, for every $\epsilon > 0$, there exists some $\delta > 0$ such that if $\|f-g\| < \delta$, then $\|Tf - Tg\| < \epsilon$.

To start off, I would just write what $\|Tf - Tg\|$ is equal to, and see where that takes me. Notice that it is equal to
$$\int_{-1}^1 |(Tf)(x) - Tg(x)|dx = \int_{-1}^0|(Tf)(x) - (Tg)(x)|dx + \int_{0}^1|(Tf)(x) - (Tg)(x)|dx$$
Now we know that in the first interval, $(Tf)(x) = f(x+1)$ and $(Tg)(x)=g(x+1)$, and similarly for the second interval, so that simplifies to
$$\int_{-1}^0|f(x+1) - g(x+1)|dx + \int_{0}^1|f(x-1) - g(x-1)|dx$$
The next steps, I leave to you, with a hint that you might want to consider some sort of change of variables in the two integrals :).
Hint 2: In the first integral, susbtitute $y=x+1$ and in the second, substitute $y=x-1$. Then, take a look at what you get and compare it to $\|Tf - Tg\|$.
