Consider $X=C[0,1]$ with its usual sup-norm topology.Let $S=\{f\in X :\int _0^1f\neq 0\}$.Is the set connected?

Question

Consider $X=C[0,1]$ with its usual sup-norm topology.Let $S=\{f\in X :\int _0^1f\neq 0\}$.Is the set connected?

I tried to conclude from the path connectedness of $S$ .But $S$ is not path connected

Answer 1

If you consider real-valued functions, the set is split is two halves by the kernel of the functional $\varphi:f\longmapsto\int_0^1f$.

More explicitly, $$ S=\varphi^{-1}(-\infty,0)\cup\varphi^{-1}(0,\infty) $$ is a disjoint union of open sets.

In the complex-valued case, $S$ is path connected.

Answer 2

This is not an answer, but wouldn't fit in a comment. It is a response to Martin.

Suppose we have $f,g \in S$. Then $z f \in S$ for all $z \in \mathbb{C} \setminus \{0\}$. Since $\mathbb{C} \setminus \{0\}$ is path connected, there is some path $\gamma$ such that $\gamma(0) = 1, \gamma(1) = {\phi(g) \over \phi(f)}$ and $\gamma$ does not pass through zero. Then the path $\gamma \cdot f$ 'changes' $f$ to $f'={\phi(g) \over \phi(f)} f$, and $\phi(f') = \phi(g)$. Now consider the path $\eta(\lambda) = (1-\lambda) f' + \lambda g$, with $\lambda \in [0,1]$. We see that $\phi(\eta(\lambda)) = \phi(g)$ for all $\lambda$ and $\eta(1) = g$.