Let $H$ be an (in general infinite dimensional) separable Hilbert space with scalar product $<\cdot,\cdot>$. Given another inner product $<\cdot,\cdot>_2$ defined everywhere on $H \times H$, will the operator $T: H \rightarrow R$ defined by $T(f) = <f,f>_2$ always be continuous (with respect to $<\cdot,\cdot>$ on $H$)?
It seems to me that there might be some version of the closed graph theorem but I did not find anything suitable. Or is it possible to give a simple counter example?
Consider the Hilbert space $\ell ^2$ of square summable sequences, with its standard basis $e_n$. Define $Te_n=ne_n$ on the subspace $\mbox{span }\{e_1,e_2,\ldots\}$. This is a densely defined, unbounded operator. Extend $E=(e_1,e_2,\ldots)$ to a Hamel Basis $F$ and put $Tf=0$ for any $f\in F\setminus E$. Then $T$ is an everywhere defined, unbounded, linear operator, and $\langle Tx,x\rangle\geq 0$. Consider now the inner product $\langle\cdot,\cdot\rangle_2=\langle (T+I) \cdot,\cdot\rangle$.
Note that this construction relies on the axiom of choice, and that T constructed above is non-closable.