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Suppose that $X$ is a Banach Space and that $T: X \to C^n [0,1]$ is a linear operator. Assume that for any $t \in [0,1]$, any $k$ with $0 \leq k \leq n$ and any sequence $(x_j)^\infty\,_{j = 1}$ in $X$ with $\lim_j \|x_j| = 0$, $\lim_j (Tx_j)^k = 0$. Show that $T$ is bounded.

Here $C^n [0,1]$ is the set of all functions $f : [0,1]\to\mathbb R$ such that $f$ has $n$ continuous derivatives, where $\|f\| = \sup_k\sup_t|f^k(t)|$.

Please help me with this question. I know that in the end, I need to show that $\|Tx\|\leq\|T\|\|x\|$ but I have no idea how to get there.

Thank you.

an4s
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taupi
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1 Answers1

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Use the closed graph theorem. $lim_j(x_j,T(x_j))=(x,y)$ implies that $lim_jx_j=x$ is equivalent to $lim_j(x_j-x)=0$ the condition says that $lim_j(T(x_j-x)^k=0$ we deduce that $lim_j(Tx_j)^k=(T(x))^k$ and $T(x)=y$ so the graph is closed.

Tsemo Aristide
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