Upper bound on an inverse operator

Question

I'm stuck on proving what seems like a rather routine estimate.

Let $H^i(\mathbb{R})$ denote the $i$-th Sobolev space. Let $T:H^2(\mathbb{R})\rightarrow H^0(\mathbb{R})$ be a bounded operator with bounded inverse $T^{-1}: H^0(\mathbb{R})\rightarrow H^2(\mathbb{R})$ such that

$$||Tu||_0\geq C||u||_0,\qquad\forall u\in H^2(\mathbb{R})$$

for a fixed constant $C$. Viewing $T^{-1}$ as a bounded operator $H^0(\mathbb{R})\rightarrow H^0(\mathbb{R})$, show that $||T^{-1}||_{B(H^0)}\leq\frac{1}{C}.$

Thanks for your help.

Answer 1

For any $v \in H^0$, the element $T^{-1} v$ belongs to $H^2$ by assumption. Applying your estimate to $u = T^{-1} v$, we get

$$ \| T^{-1} v \|_0 \leq \frac{1}{C} \| T(T^{-1} v)) \|_0 = \frac{1}{C} \| v \|_0$$

which shows that $\| T^{-1} \|_0 \leq \frac{1}{C}$.