vector space with a metric

Question

Let $C([0,1])$ be the vector space of continuous functions on $[0,1]$ equipped with the metric $\rho(f,g)= \int^1_0 \frac{|f(x)-g(x)|}{1+|f(x)-g(x)|} dx.$ Show that if $L:C([0,1])\rightarrow R$ is continuous and linear, then $L(f)=0$ for every $f\in C([0,1])$.

Answer 1

Hint:

What is $B(0,1)$ in the $\rho$ metric?

If $L$ is bounded on $B(0,1)$ then it can only be zero.

Answer 2

Some more hints: If the support of some $f\in C[0,1]$ (i.e the closure of $\{x\in[0,1]:f(x)\neq 0\}$) has measure $\le \varepsilon$ then (independently of the size of $f$) you have $\rho(f,0)\le \varepsilon$. Using a partition of unity you can write every $f$ as a sum (and hence as a convex combination) of functions with arbitrary small $\rho$-distance to $0$. The continuity of $L$ gives some $\delta$ such that $|L(f)|\le 1$ for all $\rho(f,0)<\delta$ and (pulling out convex combinations) the above implies that $|L(f)|\le 1$ holds for every $f$. This yields $L=0$.