Let $V = \mathbb{C}[X]$ and define an inner product on $V$ by setting $$ \left\langle \sum_{i=0}^{\infty}a_iX^i,\sum_{i=0}^{\infty}b_iX^i\right\rangle= \sum_{i=0}^{\infty}a_i\bar{b_i}.$$
Let $\alpha$ be the endomorphism of $V$ defined by $\alpha : p(X) \to (X+1)p(X)$. Calculate $\alpha^*$, or show that it does not exist. ($\alpha^*$ is adjoint of $\alpha$)
Let $p(X), \:q(X)\in\mathbb{C}[X]$, hence there exist scalars $a_1,\dots,a_n$ and $b_1,\dots,b_n$ in $\mathbb{C}$ satisfying $p(X)=\sum_{i=0}^{n}a_iX^i$, and $q(X)=\sum_{i=0}^{n}b_iX^i$.
Now, by the assumption we have
\begin{align} \left\langle \alpha(p(X)),q(X)\right\rangle &=\left\langle (X+1)\sum_{i=0}^{n}a_iX^i, \sum_{i=0}^{n}b_iX^i\right\rangle \\ &=\left\langle \sum_{i=0}^{n}a_iX^{i+1}+\sum_{i=0}^{n}a_iX^i,\sum_{i=0}^{n}b_iX^i \right\rangle\\ &=\left\langle \sum_{i=0}^{n}a_iX^{i+1},\:\sum_{i=0}^{n}b_iX^i\right\rangle+ \left\langle \sum_{i=0}^{n}a_iX^i,\sum_{i=0}^{n}b_iX^i\right\rangle \\ &=(0+a_0\bar b_1+a_1\bar b_2\dots +a_{n-1}\bar b_{n}+0)+(a_0\bar b_0\dots+a_n\bar b_n)\\ &=a_0(\bar b_1+\bar b_0)+\dots+a_{n-1}(\bar b_{n-1}+\bar b_n)+a_{n}(0+\bar b_n) \end{align}
I'm stuck on the rest, I'm not sure if it's true to say $$a_0(\bar b_1+\bar b_0)+\dots+a_{n-1}(\bar b_{n-1}+\bar b_n)+a_{n}(0+\bar b_n)=$$ $$\left\langle \sum_{i=0}^{n}a_iX^i,\sum_{i=0}^{n}b_iX^i\right\rangle +\left\langle \sum_{i=0}^{n}a_iX^i,\sum_{i=0}^{n}b_iX^{i+1}\right\rangle.$$ Hence, $\alpha^*:p(X)\to (X+1)p(X)$ and $\alpha $ is selfadjoint.
Note that for a polynomial $f(X)=a_0+a_1X+\dots+a_nX^n$, we have $$ \langle f(X),X^k\rangle=a_k $$ so, if $f_n(X)=\alpha^*(X^n)$, we have $$ \langle f_n(X),X^k\rangle= \langle X^n,\alpha(X^k)\rangle= \langle X^n,X^{k+1}+X^k\rangle= \begin{cases} 1 & \text{if $k=n$}\\ 1 & \text{if $k=n-1$}\\ 0 & \text{otherwise} \end{cases} $$ In particular we should have $$ \alpha^*(1)=1,\quad \alpha^*(X)=1+X\quad \dots,\quad \alpha^*(X^n)=X^{n-1}+X^n,\quad \dotsc $$
Can you go on?
If you want a representation of $\alpha*$ in terms of polynomial operations, note that $$ \alpha^*(a_0+a_1X+\dots+a_nX^n)= a_0+a_1+a_1X+a_2X+a_2X^2+\dotsb= \frac{p(X)-p(0)}{X}+p(X) $$ where the division is possible because $p(X)-p(0)$ has no constant term.