Adjoint series representation? I am aware that for a normal square matrix $M\in\Bbb C^{n\times n}$, there exists a polynomial $P$ so that $P(M)=M^*$ What if I have a normal bounded operator $T\in\mathscr L(X)$ where $X$ is a hilbert space. Does there exist a power series representation of it? ie. $T^*=\sum_{k=0}^\infty a_kT^k$ 
thanks
 A: Joint effort with Norbert. Let $T$ be the multiplication by $z$ on the Hilbert space $L^2(\mathbb T)$, where $\mathbb T=\{z:|z|=1\}$. Then $T^*$ is the multiplication by $\bar z$. Obviously $T$ and $T^*$ commute. 
For any polynomial $p$, $p(T)$ is the operator of multiplication by $p(z)$. It is well known that $\sup_{|z|=1}|p(z)-\bar z|\ge 1$ for any polynomial; this can be proved by comparing $\int_{\mathbb T} p(z)\,dz=0$ and $\int_{\mathbb T} \bar z\,dz=2\pi$. Therefore, for any polynomial $p$ we have $\|p(T)-T^*\|\ge 1$. 
By the above, a power series of $T$ cannot converge to $T^*$ in the operator norm topology. But even more can be said. Applying $p(T)$ to the function $1$, we get a complex polynomial, and the closure of complex polynomials in $L^2(\mathbb T)$ is exactly the Hardy space $H^2$. Since $\bar z\notin H^2$ (in fact, is orthogonal to it), it follows that a sequence of polynomials of $T$ cannot converge to $T^*$ even on the single vector $1$.

The answer is positive if the spectrum of $T$ has empty interior and connected complement. This follows by combining the spectral theorem with Mergelyan's theorem. Note that in the finite-dimensional case the spectrum is finite. 
