Is this is a connected subset of $\mathcal{C}[0, 1]$ as a metric space with the sup norm metric? $$K = \{f \in \mathcal{C}[0,1]| \int_0^\frac{1}{2}f - \int_\frac{1}{2}^1 f = 1\}$$
So far, I've managed to prove the following: If $f \in K$, then $f+c \in K$ for all $c \in \mathbb{R}$. If $A$ is a subset of $K$ which is both open and closed, then if $f \in A$, then $f + c \in A$ for all $c \in \mathbb{R}$.
To see this, let $$C = \{c \in \mathbb{R}|\,f+x \in A \text{ for all } x \text{ in the closed interval between } c \text{ and } 0\}$$ Suppose $C$ is bounded above, let $M$ be $\sup C$. $M > 0$ as for some $r > 0$, $B_K(f, r) \subset A$. Given any $\epsilon > 0$, there exists $c_\epsilon \in C$ with $M - \epsilon < c_\epsilon \leq M$. Thus $[0, M - \epsilon] \subset C$ for all $\epsilon > 0$, which means $[0, M)$ is a subset of $C$. It also follows that $f + M$ is a limit point of $A$, and as $A$ is closed, $f + M \in A$. So $M \in C$.
But $A$ is open, so there exists some $r > 0$ such that $B_K(f+M, r) \subset A$. Thus, $M + \frac{r}{2} \in C$, contradicting the fact that $M$ is the supremum.
Thus, $C$ is not bounded above. Analogously, we can prove that $C$ is not bounded below.
I have no idea whether this result leads to something useful or not. How do I go about this?
This solution was suggested by @saulspatz.
$K$ is path connected. If $f, g \in K$, then $tf + (1 - t)g \in K$.
$$ \int_0^\frac{1}{2} [tf - (1-t)g]- \int_\frac{1}{2}^1 [tf - (1-t)g] = t\left[\int_0^\frac{1}{2} f - \int_\frac{1}{2}^1 f\right] + (1-t)\left[\int_0^\frac{1}{2} g - \int_\frac{1}{2}^1 g\right] = 1$$
Thus, $\gamma(t) = tf + (1-t)g$ for $t \in [0, 1]$ is a path in $K$ from $f$ to $g$.
It follows that $K$ is connected.