Let $(\mathcal{H},\Vert\cdot\Vert)$ be a complex Hilbert space with inner product $\langle\cdot,\cdot\rangle$ and $A\in\mathcal{L}(\mathcal{H})$ be a bounded self-adjoint operator with spectrum $\sigma(A)\subset\mathbb{R}$.
The functional calculus is usually defined first of all for polynomials: Let $P\in\mathbb{C}[\sigma(A)]$ be a polynomial, i.e. a map $P:\sigma(A)\to\mathbb{C}$ of the type $$P(t)=\sum_{i=0}^{k}\lambda_{i}t^{i}$$ for some $k\in\mathbb{N}_{0}$ and coefficients $\lambda_{0},\dots,\lambda_{k}\in\mathbb{C}$. Then we define a bounded operator $P(A)\in\mathcal{L}(\mathcal{H})$ by $$P(A):=\sum_{i=0}^{k}\lambda_{i}A^{i}.$$
In the end, we want to define a map which assigns to each polynomial on $\sigma(A)$ a bounded operator. However, a priori it is not clear that two different polynomials result into different operators. As an example, consider the case of an operator $A$ such that the spectrum is given by $\sigma(A)=\{0,1\}$. In this case, the polynomials $P_{1}(t):=t$ and $P_{2}(t):=t^{2}$ are equivalent, but it is not clear that $P_{1}(A)=A\stackrel{?}{=}A^{2}=P_{2}(A)$.
In order to show that this is the case, one usually shows that $$\Vert P(A)\Vert_{\mathcal{L}(\mathcal{H})}=\sup_{\lambda\in\sigma(A)}\vert P(\lambda)\vert.$$ I understant how to proof this statement, by I do not see why this is enough to show that the functional calculus of polynomials is well-defined.
I would like to show that if $P_{1},P_{2}\in\mathbb{C}[\sigma(A)]$ are to polynomials such that $P_{1}(t)=P_{2}(t)$ for all $t\in\sigma(A)$, then $P_{1}(A)=P_{2}(A)$. I tried to use the paralellogram identity together with the above formula for the operator norms in order to show that $\Vert P_{1}(A)-P_{2}(A)\Vert=0$. However, I was not able to proof it.
You must be able to show that $P(A)=0$ if $P$ vanishes on the spectrum. This is not generally true if $A$ is not normal. For example, if $A$ is nilpotent of order $2$ (meaning that $A^2=0$, but $A\ne 0$,) then a polynomial $P$ that vanishes on the spectrum of $A$ does not necessarily give $P(A)=0$. However, a normal operator $A$ such that $A^2=0$ must satisfy $A=0$; more generally, $p(A)=0$ if $p$ vanishes on the spectrum of $A$, and $m(A)=0$ will also hold if $A$ is normal and if $m$ has only first order roots, but has the same roots as $p$.