Let $V$ be an infinite dimensional inner product space and $\varphi : V \rightarrow V $ a linear transformation. If the adjoint of $\varphi$ (a linear transformation $\psi: V \rightarrow V$, such that $\forall \alpha,\beta\in V: (\varphi(\alpha),\beta)= (\alpha,\psi(\beta)$) exists,then the adjoint of $\varphi$ is unique denoted by $\varphi^{\ast}$.
Considering the real vector space of polynomials $\mathbb{R}[x]$.For any $$f(x)=a_{0}+a_{1}x+\cdots+a_{n}x^{n},g(x)=b_{0}+b_{1}x+\cdots+b_{m}x^{m} \in{\mathbb{R}[x]}，$$ we define the inner product as $$(f(x)，g(x))=a_{0}b_{0}+a_{1}b_{1}+\cdots+a_{k}b_{k}，k=min\{n，m\}.$$ Thus $\mathbb{R}[x]$ be an infinite dimensional inner product space.
Question：
In the above infinite dimensional inner product space $\mathbb{R}[x]$, Let a linear transformation $$\varphi : a_{0}+a_{1}x+\cdots+a_{n}x^{n}\rightarrow a_{1}+a_{2}x+\cdots+a_{n}x^{n-1}.$$Then I can find two distinct linear transformation$$\psi_{1}:a_{0}+a_{1}x+\cdots+a_{n}x^{n}\rightarrow a_{0}x+a_{1}x^{2}+\cdots+a_{n-1}x^{n};$$$$\psi_{2}:a_{0}+a_{1}x+\cdots+a_{n}x^{n}\rightarrow a_{0}x+a_{1}x^{2}+\cdots+a_{n-1}x^{n}+a_{n}x^{n+1}.$$ Both $\psi_{1}$ and $\psi_{2}$ are the adjoint of $\varphi$, that obviously conflict the uniqueness of $\varphi^{\ast}$.What's wrong with it?
Let's test them: $$\langle\phi(x^n),x^{n-1}\rangle=\langle x^{n-1},x^{n-1}\rangle=1$$
$$\langle x^n,\psi_1(x^{n-1})\rangle=\langle x^n,0\rangle=0$$
$$\langle x^n,\psi_2(x^{n-1})\rangle=\langle x^n,x^n\rangle=1$$
There's something wrong with $\psi_1$.