is C[a,b] with 1-norm connected

Question

If C[a,b] denotes the set of all real valued continuous functions over [a,b] is it connected w.r.t. the 1-norm ?1-norm of a function f is defined to be integration of f from a to b.

Answer 1

Every normed vector space $V$ is always path-connected, hence connected.

To see this, let $x,y \in V$ be arbitrary. Then

$$ \phi : [0,1] \to V, t \mapsto (1-t) x + ty = x + t(y-x) $$

is a path from $x$ to $y$ and continuous because of

$$ \Vert \phi(t) -\phi(s)\Vert = \Vert (t-s)(y-x) \Vert = |t-s| \Vert x - y \Vert \to 0 $$

for $t \to s$.

What you should note is that the $L^1$ norm is indeed a well-defined norm on $C([a,b])$ (why?).

Answer 2

A path $t\mapsto h_t$ from $g$ to $f$ if given by $h_t(x)=tf(x)+(1-t)g(x)$, so $C[a,b]$ is even path-connected.

Note that $|h_{t+\epsilon}-h_t|_1=\epsilon\cdot |f-g|_1$ is small for small $\epsilon$.