how to prove the operator $L$ is zero. Problem
For 
$f, g \in C([0; 1])$ 
define
$$d(f,g):= \int_{0}^{1}\min\{1,\vert f(x)-g(x)\vert\}dx $$
(a) Prove that $d$ is a distance on $C([0,1]).$
(b) Let $L: C([0,1])\to \mathbb{R}$ 
be a linear operator, continuous with respect to the topology induced by 
$d$. Prove that $L = 0.$
The part (a) is easy to solve. But I have no idea how to start for part (b). Please leave some hints, not a whole solution, to me. 
 A: Using the continuity of $L$, pick $\delta > 0$ such that $|L(f)|<1$ whenever $d(f,0)<\delta$. One sufficient condition for $d(f,0)<\delta$ is that $f$ satisfy the following property:

Property $P_{\delta}$. The support of $f$ is contained in an interval of length $<\delta$. To be precise, there exists an interval $I\subseteq [0,1]$ of length $<\delta$ such that $ \{x : f(x) \neq 0 \} \subseteq I $.

If $f$ has this property, then so is any scalar multiple of $f$, hence
$$ |L(f)| = \frac{1}{n}|L(nf)| \leq \frac{1}{n} $$
and we have $L(f)=0$ by letting $n\to\infty$.
Now the key idea is that any continuous function $f$ can be uniformly approximated by linear combinations of functions having poperty $P_{\delta}$.
For instance, for each $n$ satisfying $n>2/\delta$, we notice that the linear interpolation $f_n$ of points $(\frac{k}{n}, f(\frac{k}{n}))$ for $k = 0, \cdots, n$ can be written by
$$f_n(x) = \sum_{k=0}^{n} f\left(\frac{k}{n}\right)\varphi_{n,k}(x), \qquad \varphi_{n,k}(x) := \max\left\{0, 1-n\left|x-\frac{k}{n}\right|\right\}.$$
Since each $\varphi_{n,k}$ is supported on the interval $[\frac{k-1}{n}, \frac{k+1}{n}]$ whose length is $=\frac{2}{n} < \delta$, it has property $P_{\delta}$ and hence $L(\varphi_{n,k})=0$. So it follows that $L(f_n) = 0$. On the other hand, we can easily check that $f_n \to f$ uniformly and hence $f_n \to f$ under the metric $d$. So we have $L(f)=0$ as desired.
