Linear map $L : C([0,1])\rightarrow C([0,1])$ continuous? Let $X=C([0,1])$ equipped with the norm $\Vert\cdot \Vert=\max_{x\in [0,1]}|f(x)|$. If $L:X\rightarrow X$  is linear, is $L$  continuous?
If not, what if $Lf\ge 0$ for $f\ge0$ for $\forall x \in [0,1]$ is assumed?
Edit: Rephrased the question.
Edit2: Attempt,
Let, $\Vert f_n \Vert \rightarrow 0$ when $n\rightarrow \infty$ and $f_n=\sum_{i=0}^{\infty}\alpha^{(n)}_i x^{i} $ then $\Vert Lf_n\Vert=\max_{x\in [0,1]}|\sum_{i=0}^{\infty}\alpha^{(n)}_i Lx^{i}| \rightarrow 0$ Because $\forall \alpha^{(n)}_i \rightarrow 0$ when $n\rightarrow \infty$.
But i don't know if this is allowed?
 A: Let $X$ be a normed vector space and $Y$ be a normed vector space, $Y \neq \{ 0 \}$. If $X$ is infinite dimensional and incomplete, then there exist discontinuous linear maps $L : X \to Y$. Often, these can be explicitly constructed. For instance, if $X=C^1$ with the sup norm and $Y=C^0$ with the sup norm, then $L(f)=f'$ is a discontinuous linear map, since $f_n=\frac{1}{n} \sin(n^2x)$ converges in $C^1$ but $L(f_n)$ does not converge in $C^0$.
If $X$ is infinite dimensional and complete, then one can use the axiom of choice to nonconstructively prove that there exist discontinuous linear maps $L : X \to Y$. $C([0,1])$ with the sup norm is infinite dimensional and complete, so this case applies to your question.
A: Show that $\Vert Lf\Vert\le\alpha$ for all $\Vert f \Vert=1$. 
Because 
$1\ge \pm f, \forall x \in [0,1]$ 
we have that 
$1\pm f\ge 0$, 
thus, 
$L(1\pm f)\ge 0$. 
By linearity 
$-L1\le Lf\le L1$, 
that is 
$|Lf|\le L1, \forall x \in [0,1]$, 
hence 
$\max_{x\in[0,1]}|Lf|= \Vert Lf \Vert \le \max_{x\in [0,1]}L1=\alpha$. 
Because boundness gives continuity the conclusion follows.
