Suppose $L: H \to H$ is a continuous linear transformation with operator norm $\|L\|$ and $c > \|L\|$.
show that $cI - L$ is continuous and find a bound on its norm
Show that $cI - L$ has closed range.
Prove that $cI - L$ has a continuous inverse operator and find its bound
My idea for the first part is that $cI$ is the same as just multiplying an element of $H$ by the constant $c$ which is finite so this should be a bounded linear operator, and therefore a continuous linear operator. Then, $cI - L$ is just a difference of linear operators which is again a continuous linear operator. The bound for $cI$ should be $c$, I think, and since $c > \|L\|$ we should be able to claim that $c > \|cI -L\|$
For the second part, I know a theorem that says: if $L: H \to H$ is a continuous linear transformation and there is a $k > 0$ s.t. $\|Lu\| \ge k\|u\|$ for all $u$ then $L$ is one-to-one and has closed range. I want to apply this to my solution, but I'm having trouble thinking of a way to show that I have such a $k$.
I have no clue about the last part
Since $cI$ is continuous with bound $c$ it follows that $cI-H$ is continuous.
The bound $\|(cI-H)x\| \le \|cx\|+\|Hx\| \le (c+L)\|x\|$, hence $\|cI-H\| \le c+L$.
Note that $\|(cI-H)x\| \ge \|cx\|-\|Hx\| \ge (c-L) \|x\|$.
Let $G = {1 \over c} H$, and note that $\|G\| < 1$. Define $F = \sum_{k=0}^\infty G^k$ and note that $\|F\| \le \sum_{k=0}^\infty \|G\|^k = {1 \over 1-\|G\|}$ so $F$ is bounded.
Now note that $FG = GF = I$, so $F$ is the inverse of $G$, hence ${1 \over c}F$ is the inverse of $H$.
A bound is easy to compute from the bound on $F,G$.