Given that a linear operator $T:X\rightarrow Y$, where $X$ and $Y$ are both Banach spaces, $D$ a dense subspace of $X$, if we know that the restriction of $T$ to $D$, say, $S=T|_{D}$ is bounded, then it is a famous theorem in functional analysis that $S$ has a unique bounded extension $\widetilde{S}$ on $X$, now is it always necessary that $\widetilde{S}=T$? So now the equivalent question is, I am looking for some unbounded linear operator which its restriction to some dense subspace is bounded.

The second question is similar, does the operator norm $\|T\|$ always satisfy that $\|T\|\leq\|S\|$?

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    $\begingroup$ No. Just start out with a bounded operator on $D$ and make $T$ unbounded on the whole space (recall that you can assign arbitrary values on a basis). More precisely, you will have $\widetilde{S}=T$ if and only if $T$ is closed. $\endgroup$ – user138530 Mar 8 '16 at 4:59

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