The existance of linear isomorphism Let, $(E,\| \cdot \|)$ be a normed linear space and $\phi$ a linear functional on $E$ which is not continuous. Let, $y_0\in E$ such that $\phi(y_0)=1$. Define $S:E\to E$ such that $S(x)=x-2\phi(x)y_0$. Define a new norm on $E$ by $\|x\|_{\phi}:= \|S(x)\|$. Then show that
(a)$S^2=I$, where $I$ is the identity operator on $E$.
(b)There is a linear isomorphism $T:E\to E$ such that $\|T(x)\|_{\phi}=\|x\|\ \forall x\in E$
(c)Show that $\| \cdot \|_{\phi}$ is not equivalent to $\| \cdot \|$
My attempts:
(a) Since we have $S(x)=x-2\phi(x)y_0$, so, $SS(x)=S(x)-2\phi(x)S(y_0)$ so, $S^2(x)=S(x)-2\phi(x)S(y_0)$. Now, $S(y_0)=y_0-2\phi(y_0)y_0$, so, $S(y_0)=-y_0$, So we get $S^2(x)=x \ \forall x$. So, we are done.
(c) If possible let the two norms are equivalent then then convergence in $\| \cdot \|$ implies convergence in $\| \cdot \|_\phi$. Let $x_n\to x$ in $\| \cdot \|$ but, we can to say any thing about $S(x)$ because $\phi$ is not continuous. So contradiction.
But I can not understand how to proceed (b)? Any hints will be very much helpful.
Thanks in advance.
 A: The answer to (b) depends a little on what definition of isomorphism you use. Is it an isomorphism of vector spaces, i.e. an invertible linear map? Or is it an isomorphism of normed linear spaces, i.e. a bounded invertible linear map whose inverse is also bounded? The latter would flatly contradict (c), as it would make the topologies generated by the two norms the same, so we are just left with the former.
Assuming we are looking for the former, then the obvious choice from part (a) is $T = S$. In this case,
$$\|Tx\|_\phi = \|Sx\|_\phi = \|S(Sx)\| = \|S^2x\| = \|Ix\| = \|x\|.$$
Note that $S$ is a linear operator and $S$ is its own inverse, hence it is invertible. This gives you the linear isomorphism you're looking for.
I have a comment too about your approach to part (c). Writing "we can't say anything" is, to me, a weak conclusion. You should either show a counterexample exists or, possibly more elegantly, show that if $\| \cdot \|$ and $\| \cdot \|_\phi$ are equivalent, then $S$ must be bounded, and hence $\phi$ must be bounded. The contrapositive of this statement is what you want proven.
