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Suppose that X={f $\in$ C([$0,1$]): $\exists \delta>0$ f|[$0,\delta$]=$0$}, $||f||=sup${$|f(t)|: t \in [0,1]$} and T: X -> X, (Tf)(x)=$\frac{1}{x}$f(x), for $x \in [0,1]$. Prove that T has closed graph but is not continuous.

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  • $\begingroup$ There is a problem when you define T because 1/x f(x) is not define on x=0 $\endgroup$ – Federico Fallucca May 14 at 13:19
  • $\begingroup$ @FedericoFallucca: That is not an issue because of the way $X$ is defined. $\endgroup$ – copper.hat May 14 at 13:25
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Let $f_n(x) = \min({2 \over n}, \max(0, 2(x-{1 \over n})))$. Note that $f_n \to 0$ but $\|Tf_n\| = 1$ for all $n$ (since $f_n({2 \over n}) = {2 \over n}$ for $n \ge 2$). Hence $T$ is not continuous.

Suppose $f_n \to f$ and $Tf_n \to y$. We want to show that $Tf = y$. Note that $y \in X$ by assumption.

For $x>0$ we see that ${1 \over x} f_n(x) \to y(x)$ and so ${1 \over x} f(x) = y(x)$.

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