Mathematics Asked on January 3, 2022

Assume $T_n,T$ are bounded bijective linear operators $X to Y$ and $T_n to T$ pointwise. Show $T_n^{-1}to T^{-1}$ pointwise $iff$ $|T_n^{-1}|leq C$

Note: $X,Y$ are banach spaces.

My proof:

Forward direction is uniform boundedness principle. Backwards:

Let us assume that $T_n^{-1}(y)not to T^{-1}(y)$ so there is a subsequence and $y$ s.t $|T_{n_k}^{-1}(y)- T^{-1}(y)|geq epsilon$. Now we know that $T_{n_k}(x)to T(x)$ and since $T^{-1}_{n}$ are uniformly boundaed we get that $|T^{-1}_{n_k}(T_{n_k}(x)-T(x))|<C|T_{n_k}(x)-T(x)|to 0$ Now just let $x=T^{-1}(y)$ and we arrive at contradiction. Is this correct? It seems roundabout in my opinon and there is probably more direct way to do it.

You can do it directly without the contradiction. Let $yin Y$ be arbitrary. Since ${T_n}, T$ are assumed to be bijective functions, there exist $T^{-1}(y) = xin X$. We also have a sequence of vectors $T^{-1}_n(y) = x_n$. You noticed that $T^{-1}_n(T_n(x)) = x$ and $T^{-1}_n(T(x)) = T^{-1}_n(y) = x_n$. Hence $$begin{align*}|| T_n^{-1}(y) - T^{-1}(y) || = ||x_n - x || &= ||T^{-1}_n(T(x)) - T^{-1}_n(T_n(x)) ||\ &= ||T_n^{-1} (T(x) - T_n(x))||\ &leq C ||T(x) -T_n(x)||to 0 end{align*}$$ Since $T_nto T$ pointwise.

Answered by Andrew Shedlock on January 3, 2022

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