Prove that linear operator is bounded Let $f$ be a continuous and bounded function $f\in C(X)\cap B(X)$, $L_f$ is a linear operator,  $L_f: g\rightarrow fg$, $g \in C(X) \cap B(X)$. I try to prove that if $\lambda\neq f(x),\forall x\in X$, then   $(L_f -  \lambda I)^{-1}$ is bounded.
I know that $||A|| = \sup_{||x|| \leq 1}||A_x||$ and $A$ is bounded if $||A|| < \infty$. But since i'm rather new to such problems i don't understand what is the norm of $C(X)\cap B(X)$ and how to use it. Any pieces of advice will be useful.
 A: This is not true in general. Consider $X = (0, 1)$ and $f$ the identity on $X$. Then $f$ is bounded and continuous and $\lambda = 0$ does not lie in the image. However, $x \mapsto \frac{1}{x}$ is not bounded. 
If we impose the condition on $X$ that it is compact, then this should work out, as $f$ will have a closed image. 
If $\lambda \notin f(X)$, there is a $\varepsilon > 0$ such that $B_\varepsilon(\lambda) \cap f(X) = \emptyset$. 
This gives: $\frac{1}{\vert f(x) - \lambda \vert} \leq \frac{1}{\varepsilon}$. So in operator-language: Since $(L_f - \lambda \cdot 1)^{-1}$ is multiplication by $h := \frac{1}{f(x) - \lambda}$ (as $h \in C_b(X)$ and by uniqueness of the inverse), we have $$\Vert (L_f - \lambda \cdot 1 )^{-1} \Vert = \sup_{\Vert g \Vert_\infty \leq 1} \Vert h \cdot g \Vert_\infty \leq \Vert h \Vert_\infty \leq \frac{1}{\varepsilon}$$
So the operator is bounded. 
However, this might be a bit too strong an assumption, as $C(X) = C_b(X)$ for any compact $X$. 
A weaker assumption that would suffice, is that $\lambda$ does not lie in the spectrum of $L_f$ (which is necessary anyway for $(\lambda \cdot 1 - L_f)^{-1}$ to be well-defined). As Daniel Fischer pointed out in the comments, the spectrum is closed and since $f(X) \subset \sigma(L_f)$, we have a $\varepsilon > 0$ such that $B_\varepsilon(\lambda) \cap f(X) = \emptyset$. 
The image of $f$ and the spectrum of $L_f$ correspond if $X$ is compact (as for $\lambda \notin \mathrm{cl}(f(X))$ we can find an inverse). 
