Let $f$ be a function defined on $\ell^{2}$ to $\ell^1$ by $f(x) =(x_n/ n)$. Show $f$ is uniformly continuous.
My attempt at the proof: Let $\{a_n\}$ and $\{b_n\}$ be sequences in $\ell^2$ Let $\varepsilon > 0$, choose $\delta = 6\varepsilon/\pi^2$. Then $d(f((a_n),(b_n)))= \left(\sum_{n=0}^\infty ({a_n-b_n\over n})^2\right)^{0.5} \le \left(\sum_{n=0}^\infty {(a_n-b_n)^2\over n^2}\right)^{0.5} \le \left(\sum_{n=0}^\infty {(a_n-b_n)^2}\right)^{0.5} \left(\sum_{n=0}^\infty ({1 \over n^2})\right)^{0.5}$... After some more typing, I end up with $|a_n-b_n|\cdot\delta$, which I then sub delta and get $\varepsilon$.
Does this show that f is uniformly continuous?
Hint: If $X,Y$ are normed linear spaces and $T:X\to Y$ is a bounded linear map, then $T$ is not just uniformly continuous, it is Lipschitz, with Lipschitz constant equal to $\|T\|,$ the operator norm of $T.$ So we only need to show $f: l^2 \to l^1$ is a bounded linear operator. Wait! We don't even know yet that $\{a_n\}\in l^2 \implies f(\{a_n\})\in l^1.$ There is a very well known inequality that shows that. Once you have that, the fact that $f$ is bounded and linear is simple.