The map $\phi : \ell^{1}(\mathbb{N}) \to L^{1}(\mathbb{R})$ where $\left\{ a_n \right\} \mapsto \sum_n a_n 1_{[n,n+1]}$ is norm preserving?

Question

I'm actually just trying to show $\phi$ is continuous by showing that it is Lipschitz. However, I believe the map $\phi : \ell^{1} \to L^{1}(\mathbb{R})$ where $\left\{ a_n \right\} \mapsto \sum_n a_n 1_{[n,n+1]}$ may also be norm-preserving. Here is what I have so far:

Take $x = \left\{ a_n \right\} \in \ell^{1}$ and $f = \sum_n a_n 1_{[n,n+1]} \in L^{1}(\mathbb{R})$ so that

\begin{align*} \|\ \phi(x) \|_{L^{1}(\mathbb{R})} &= \|\ \sum_{n} a_n 1_{[n,n+1]} \|_{L^{1}(\mathbb{R})} \\ &= \|\ f \|_{L^{1}(\mathbb{R})} \\ &= \int_{\mathbb{R}} |f(x)| dx \\ &= \sum_{n=1}^{\infty} \left( \int_{-n}^{-n+1} |f(x)|dx + \int_{n-1}^{n} |f(x)|dx \right) \end{align*}

Answer 1

Indeed, the map is norm-preserving (and therefore Lipschitz). All that remains is the observation that $\int_{n-1}^{n} |f(x)|dx = |a_{n-1}|$ for integers $n \geq 1$ (or $n \geq 2$, depending on your $\Bbb N$) and is $0$ for other $n$.

It may seem odd that question might ask you to prove a weaker result instead of the strongest result possible, but this makes for a tougher problem. If they had said off the bat that the map is an isometry (or provided you with a reasonable Lipschitz constant), it would take less insight to prove that the function is Lipschitz after all.