I recently learned that Banach spaces are barrelled, i.e any convex, balanced, absorbing and closed subset is a neighborhood of zero (wikipedia). I’m having trouble understanding why the following example fails.

The space $C((0,1))$ of real-valued (uniformly) continuous functions over the compact $(0,1)$ is a Banach for the sup norm. Fix $M$ and $L>0$. Let $X$ the subset consisting of $L$-lipschitz and $M$-bounded functions.

On the one hand, I checked that (and evidently made a mistake somewhere)

- $X$ is convex: since convex combination of $L$-lipschitz is $L$-lipschitz, and convex combination of $M$-bounded is $M$-bounded
- $X$ is balanced: since $-X = X$
- $X$ is absorbing: for $f$ (uniformly) continuous over $(0,1)$, shrinking $f$ makes it more and more lipschitz, and more and more bounded, so $lambda f in X$ for some $lambda>0$
- $X$ is closed: since uniform limit of $L$-lipschitz is $L$-lipschitz, and uniform limit of $M$-bounded is $M$-bounded

On the other hand, $X$ is compact (because bounded and equicontinuous, by Arzela-Ascoli theorem), so it cannot contain an open ball (open balls of $C((0,1))$ are non-compact by Riesz theorem). So it cannot be a neighborhood of zero.

So $X$ is a barrelled subset of a Banach space, yet it cannot be a neighborhood of zero… Where is the mistake?