Calculating distance between element and subspace

Question

Let $L=\{(\xi_j)\in \ell_1 \ \colon \sum_{j=1}^\infty \frac{j}{j+1}\xi_j=0\}$, which is a subspace in $\ell_1$. I want to find $$d(e_1,L)=\inf\{\|e_1-y\|\ \colon y\in L\},$$ where $e_1=(1,0,0,\ldots)$. Intuition says that it should be $\frac12$, but any ideas on how to prove it rigorously?

Answer 1

Consider $f = (\frac{1}{2},\frac{2}{3},\frac{3}{4},...) \in \ell_\infty$. Since $L = \ker f$, and by definition of operator norms, we have for all $y \in L$ that $||f||_\infty ||e_1 - y||_1 \geq$ $|f(e_1 - y)| = \frac{1}{2}$. Since $||f||_\infty = 1$, this implies $dist(e_1,L) \geq \frac{1}{2}$.

To see that in fact $dist(e_1,L) = \frac{1}{2}$ consider the sequence $y_n = e_1 -\frac{n+1}{2n}e_n$: We have $y_n \in L$ for all $n$ and $\lim_{n \rightarrow \infty} ||e_1 - y_n||_1 = \frac{1}{2}$.