where,
I know that $||T||\leq 1$ but I cant seem to show the it is larger than 1. I can't seem to come up with any functions that would show that $||T||\geq 1$.
Any ideas?
Secondly, I'm assuming that inverse of T does not exist since it is not injective as $L(f)=L(f+c)$ where c is a constant
With the given norm, the operator is continuous with norm at most 1. Indeed, $$ ||Tf||_\infty=\sup_{x\in [0,1]}|f'(x)|\leq \sup_{x\in [0,1]}(|f(x)|+|f'(x)|)=||f|| $$ As you say, the operator cannot be invertible as constants are in the kernel, and so it is not injective.
To compute the norm, take $f_n(x)=x^n$. Then, $$ ||x^n||=1+n $$ since $$ \sup_{x\in [0,1]}(|x^n|+|nx^{n-1}|)\leq ||x^n||_\infty+||nx^{n-1}||_\infty=1+n $$ but $$ \sup_{x\in [0,1]}(|x^n|+|nx^{n-1}|)\geq 1^n+n1^{n-1} $$ and $$ ||nx^{n-1}||_\infty=n $$ so $$ \frac{||Tf_n||_\infty}{||f_n||}\to 1 $$