Triangle Inequality considering Inverse Operator

Question

Let $E,F$ be two normed spaces and $T,S \in L(E,F)$ be invertable and also let $(T+S)$ be boundedly invertable. Can we assume that $||(T+S)^{-1}|| \le ||T^{-1}|| + ||S^{-1}||$ holds?

Answer 1

I've just solved my own query in the comments: The relation we have for all bounded, invertible operators in general is that $\|T^{-1}\| \geq \frac{1}{\| T\|}$. This being the case we have

$$\|(T+S)^{-1}\| \geq \frac{1}{\| T +S\|} \geq \frac{1}{\|T \| + \|S\|}$$

and this is probably as much as we could say about it without more information about specific operators. You may be able to manipulate this into some you could use, however, I'm certain the statement in your question doesn't hold in general.

Consider the multiplication operators $(\mathbb{I}f)(x) = f(x)$ and $(-\frac{3}{2}\mathbb{I}f)(x) = -\frac{3}{2}f(x)$ for all functions $f$ in the space. Then these operators are bounded and invertible, hence looking to the left hand side of your proposed inequality we have

$$\| (\mathbb{I} - \frac{3}{2}\mathbb{I})^{-1}\| = \|(-\frac{1}{2}\mathbb{I})^{-1}\| = \|-2\mathbb{I}\| = 2.$$

Similarly for the right hand side:

$$\|(\mathbb{I})^{-1}\| + \|(-\frac{3}{2}\mathbb{I})^{-1}\| = \|\mathbb{I}\| + \|-\frac{2}{3}\mathbb{I} \| = 1 + \frac{2}{3} = \frac{5}{3}.$$

Hence we can see that LHS$>$RHS.