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Exercise :

Let $X$ be a Banach space, $T in B(X)$ which means that $T$ is a bounded linear operator $T : X to X$. We suppose that for all $y in X$ the series
$$sum_{n=1}^infty|T^ny|$$
converges. Show that for all $y in X$, the equation $x = y + Tx$ has a unique solution.

Attempt/Thoughts :

Since $X$ is Banach then the space $B(X)$ of all the bounded operators from the Banach space $X$ to $X$, is also Banach. This means that $T_n to T$ as $nto infty$ over $B(X)$.

Since the given series is convergent, this means that :

$$Bigg|sum_{n=k+1}^infty|T^ny|Bigg| < epsilon $$

Also, since $T$ is a bounded operator, this means that :

$$|Ty| leq M |y|$$
I really do now know how to use the given facts to yield a result. Any given hints or elaborations will be greatly appreciated !

Formally, $x$ would be given by
$$ x "=" (1-T)^{-1} y$$
where by the formula for a geometric series, we should have
$$ (1-T)^{-1} "=" sum_{n=0}^infty T^n $$The content of the exercise is to prove that this makes sense. If we let $x_N := sum_{n=0}^N T^n y $ then the given assumption implies that $x_N$ converges in $X$ to some $x:=sum_0^infty T^n y$, and moreover
$$(1-T)^{-1}y := sum_{n=0}^infty T^n y$$
is a bounded map in $B(X)$. Then just compute (since $Tin B(X)$)
$$ (1-T) x = lim_{Ntoinfty }(1-T) x_N = lim_{Ntoinfty} y - T^{N+1}y = y$$
Its also the case (with a similar proof) that for any $y$ of the form $y=(1-T)x$,
$$(1-T)^{-1} y = x.$$
This proves that $1-T$ is invertible, and uniqueness follows.