Can someone help me solve the following please? Thanks!
Let $T: P_1(\mathbb{R}) \to P_1(\mathbb{R})$ be defined as $T(f)(x) = f'(x)+f(x).$ Determine $R(T)$ and $N(T)$. Consider $P_1(\mathbb{R})$ as an inner product space with inner product $$\langle f, g \rangle = \int_{-1}^{1} f(x)g(x)dx.$$ Evaluate $T^*$ at the vector $h(x) = 1+x$.
If $f\in P_{1}(\textbf{R})$, then $f(x) = ax + b$. Thus $T(f)(x) = a + b + ax$.
We may now determine the kernel and image of $T$. Let's start with the kernel. One has \begin{align*} T(f)(x) = a + b + ax = 0 \Longrightarrow a = b = 0 \Longrightarrow \ker(T) = \{0\} \end{align*}
Based on the rank-nullity theorem, we conclude that $\text{Im}(T) = P_{1}(\textbf{R})$.
We may now approach the second part.
Firstly, let us reinforce that the adjoint operator satisfies $\langle T(f)(x), h(x)\rangle = \langle f(x), T^{*}(h)(x)\rangle$, which is unique for each $h(x)$.
Then we need to determine the next integral \begin{align*} \int_{-1}^{1}(ax + a + b)(x+1)\mathrm{d}x = \int_{-1}^{1}(ax^{2} + (2a+b)x + a + b)\mathrm{d}x = \frac{8a}{3} + 2b \end{align*}
In order to find $T^{*}(h)(x)$, let us assume it equals $cx + d$. Consequently, \begin{align*} \int_{-1}^{1}(ax+b)(cx + d)\mathrm{d}x = \int_{-1}^{1}(acx^{2} + (ad+bc)x + bd)\mathrm{d}x = \frac{2ac}{3} + 2bd \end{align*} Comparing both expressions, we conclude that $c = 4$ and $d = 1$.
Finally, we obtain that $T^{*}(h)(x) = 4x + 1$, and we are done.
Hopefully this helps.