Show that a subset of $C(\mathbb{R})$ is compact with norm $\lVert u\rVert = \sup\lvert\frac{u(x)}{x^2 +1} \rvert$ Let $X = \{u \in C(\mathbb{R}): \lvert u(x) - u(y)\rvert \leq \lvert x - y\rvert\}$ and $C = \{u \in X: u(0) = 0\}$.
I have to prove that $C$ is compact with the norm $\lVert u\rVert = \sup_{x\in\mathbb{R}}\lvert\frac{u(x)}{x^2 +1} \rvert$.
Now, the functions in X are equi-lipschitz so they are equi-continuous and the members of C are bounded by $\lvert x \rvert$ since $u(0) = 0$ which gives $\lVert u \rVert \leq \frac12$ for all $u \in C$, so they are also uniformly bounded with this norm.
This leads me to suspect that I have to apply the Ascoli-Arzelà theorem in some form, but obviously $\mathbb{R}$ is not compact and $C$ is not uniformly bounded with the sup-norm.
I also tried to verify directly the definition of compactness but that does not seem to lead anywhere, so the best I managed to do is prove that C is closed and bounded.
I'm stuck here, any help would be greatly appreciated.
 A: The two key steps here are a diagonalizing argument and an estimation of the norm given in terms of the sup norm, $|| \cdot ||_{\infty}$, for functions in $X$.
First observe that by the definition of $C$, if $f \in C$, for all $x \in \mathbb{R}, |f(x)| \leq |x|$. For $f \in C([-n, n]),$ define $$ ||f||_n = \sup_{x \in [-n, n]} \frac{f(x)}{ x^2 + 1}.$$ Then $||\cdot ||_n$ is a norm on $C([-n, n])$, for any $f \in C([-n, n])$, $||f||_n \leq ||f||_\infty$, and by our estimation on elements of $C$ for any $f \in C$, $$ 0 \leq ||f|| - ||f|_{[-n, n]} ||_n \leq \frac{1}{n}.$$
Inductively form a sequence of subsequences as follows. Let $f_{0,k} = f_k$. Suppose $f_{n-1, k}$ is defined. Then then $[-n, n]$ is compact, and as you've shown the sequence $f_k$ is equicontinuous and uniformly bounded on $[-n, n]$. So Arzela-Ascoli gives a convergent subsequence $f_{n, k}$ of $f_{n-1, k}$ with respect to the sup norm on $[-n, n]$, hence with respect to $||\cdot ||_{n}$ on $[-n, n]$.
Define now our subsequence of interest as $g_{k} = f_{k, k}$. Our construction guarantees that $g_k$ has a continuous limit $g$ so that $g_k$ converges to $g$ uniformly on every subset of the form $[-n, n]$ hence uniformly on compact subsets of $\mathbb{R}$ and that $\lim_{n \to \infty} ||g_k - g||_n = 0$ for all $n$. So let $\epsilon > 0$, take $n > 2/\epsilon$, and then take $k$ so that for all $k' > k$, $||g_{k'} - g_k||_n < \epsilon/2$. Using our estimate above, we see that for all $k' > k$, $$||g_{k'} - g_k|| \leq \frac{1}{n} + ||g_{k'} - g_k||_n < \frac{\epsilon}{2} + \frac{\epsilon}{2} = \epsilon.$$
