I am a little bit lost and not sure if I have understood everything correctly. Maybe you can give me feedback on my subsequent proof.
The question is:
Let $c$ be the space of convergent sequences equipped with the norm $\Vert(x_k)\Vert_\infty = \sup \{\,\vert x_k\vert, k \in \mathbb{N}\,\}$.
Show that the function $\lim\colon c \rightarrow \mathbb{R}$, that maps $(x_k) \mapsto \lim_{k \to\infty} x_k$ is not only continuous but also Lipschitz-continuous.
My proof:
Be $(x_k) \in c$ and $(x_{kn})$ a convergent sequence of elements of $c$ with the the limit $\lim_{n \to \infty}(x_{kn})= (a_k )$. So there exists a $n_0 \in \mathbb{N}$ for a given $\delta >0$ so that for all $n>n_0$: $\Vert(x_{kn})-(a_k)\Vert_\infty = \sup\{\,\vert x_{kn}-a_k\vert, k \in \mathbb{N}\,\} < \delta$.
Let for an $\epsilon >0$ be: $\vert \lim(x_{kn})- \lim(a_k)\vert=\vert\lim_{k \to \infty}x_{kn}- \lim_{k \to \infty}a_k\vert < \epsilon$. As $(x_{kn})$ is convergent for all $n$ and $(a_k)$ is a convergent sequence I can write: $\vert \lim_{k \to \infty}(x_{kn} -a_k )\vert<\epsilon$. Then we conclude for all $n>n_0$: $\vert\lim_{k \to \infty}(x_{kn} -a_k)\vert \leq \sup \{\,\vert x_{kn}-a_k\vert, k \in \mathbb{N}\,\} = \Vert(x_{kn})-(a_k)\Vert_\infty < \delta$.
If we set $\delta := \epsilon$, it yields a $\delta$ which fulfils the $\epsilon$-$\delta$-criteria for continuity. With $L = 1$, the function $\lim\colon c \rightarrow \mathbb{R}$ is also Lipschitz-continuous. $\Box$
Is this correct? Or do you have any suggestions?
best regards, Philipp
I do not quite understand your proof. But the fact itself is easy to understand: the limits of two convergent sequences cannot differ by more than the maximum distance between the sequences. I would argue by contradiction: suppose $\lim x_k=l$, $\lim y_k=m$. If $|l-m|>\sup |x_k-y_k|$ we would have $|x_n-y_n|>\sup |x_k-y_k|$ for large $n$ (since $x_k-y_k\to l-m$), and this contradicts the definition of $\sup$.