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Suppose that the inner product on $P_2(\mathbb{R})$ is defined by $$\langle f,g \rangle:= f(-1)g(-1)+f(0)g(0)+f(1)g(1).$$

Consider the operator $T \in B(P_2(\mathbb{R}))$ which is defined as $Tf=f'$, so the derivative of $f$. Find the adjoint of $T$.

I am trying to find the adjoint in the following way. Since $\langle Tf, g\rangle = \langle f, T^*g\rangle$ then we have that $\langle f, T^*g\rangle = f'(-1)g(-1) + f'(0)g(0) + f'(1)g(1)$. How do I continue now since I don't know how $T^*$ looks like? I appreciate your help.

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I don't think you can get it directly. But you can do the following. Consider an orthonormal basis. For instance, $$\tag1 \frac{1}{\sqrt3},\ \frac{x}{\sqrt2},\ \frac{\sqrt3}{\sqrt2}\,(x^2-\frac23). $$ We have $$ T(\frac1{\sqrt3})=0,\ \ T(\frac{x}{\sqrt2})=\frac1{\sqrt2}=\frac{\sqrt3}{\sqrt2}\,\frac1{\sqrt3},\ T(\frac{\sqrt3}{\sqrt2}\,(x^2-2/3))=\frac{\sqrt3}{\sqrt2}(2x)=2\sqrt3\,\frac{x}{\sqrt2}. $$ So the matrix of $T$ with respect to the orthonormal basis $(1)$ is $$ T=\begin{bmatrix} 0&\sqrt3/\sqrt2&0\\ 0&0&2\sqrt3\\ 0&0&0\end{bmatrix}, $$ and so $T^*$ has matrix $$ T^*=\begin{bmatrix} 0&0&0\\ \sqrt3/\sqrt2&0&0\\ 0&2\sqrt3&0\end{bmatrix}. $$ This tells us that $$ T^*1=\sqrt3\,T^*(\frac1{\sqrt3})=\sqrt3\,\frac{\sqrt3}{\sqrt2}\,\frac{x}{\sqrt2}=\frac{3}2\,x, $$ $$ T^*x=\sqrt2\,T^*(\frac{x}{\sqrt2})=\sqrt2\,\,2\sqrt3\,\frac{\sqrt3}{\sqrt2}(x^2-2/3)=6x^2-4, $$ $$ T^*x^2=T^*(x^2-\frac23)+T^*(\frac23)=T^*(\frac23)=\frac23\,T^*1=\frac23\,\frac{3}2\,x={x}{}. $$ In summary, $$ T^*(ax^2+bx+c)={ax}{}+b(6x^2-4)+\frac{3 c x}2 =6bx^2+\left(a{}+\frac{3 c}2 \right)x-4b. $$

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  • $\begingroup$ why is $T^{*} 1=T^{*} (\frac{1}{\sqrt{3}})$? Shouldn't it be $T^{*}=\sqrt{3}T^{*}(\frac{1}{\sqrt{3}})$ $\endgroup$ – mandella Feb 10 at 9:20
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    $\begingroup$ Yes, of course. I'll edit a bit later when I have time. $\endgroup$ – Martin Argerami Feb 10 at 13:44
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    $\begingroup$ I have edited it now. Hopefully it's right. $\endgroup$ – Martin Argerami Feb 10 at 15:22
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Every $f\in P_2$ can be written as $$ f=f(-1)\frac{1}{2}x(x-1)+f(0)(1-x^2)+f(1)\frac{1}{2}x(x+1)$$ Therefore, $$ f'(x)=f(-1)(x-\frac{1}{2})-2f(0)x+f(1)(x+\frac{1}{2}) $$ which gives \begin{align} f'(-1)&=-\frac{3}{2}f(-1)+2f(0)-\frac{1}{2}f(1) \\ f'(0)&=-\frac{1}{2}f(-1)+\frac{1}{2}f(1) \\ f'(1)&=\frac{1}{2}f(-1)-2f(0)+\frac{3}{2}f(1) \end{align} So $$ \left[\begin{array}{c}(Tf)(-1)\\ (Tf)(0)\\ (Tf)(1)\end{array}\right] = \left[\begin{array}{ccc}-\frac{3}{2} & 2 & -\frac{1}{2} \\ -\frac{1}{2} & 0 & \frac{1}{2} \\ \frac{1}{2} & -2 & \frac{3}{2}\end{array}\right]\left[\begin{array}{c} f(-1)\\f(0)\\f(1)\end{array}\right] $$ Therefore, $T^*$ is represented by the transpose $$ \left[\begin{array}{c}(T^*f)(-1)\\ (T^*f)(0)\\ (T^*f)(1)\end{array}\right] = \left[\begin{array}{ccc}-\frac{3}{2} & -\frac{1}{2} & \frac{1}{2} \\ 2 & 0 & -2 \\ -\frac{1}{2} & \frac{1}{2} & \frac{3}{2}\end{array}\right]\left[\begin{array}{c} f(-1)\\f(0)\\f(1)\end{array}\right] $$ Therefore, \begin{align} T^*f &= (T^*f)(-1)\frac{1}{2}x(x-1) \\ & +(T^*f)(0)(1-x^2)\\ &+(T^*f)(1)\frac{1}{2}x(x+1) \\ &= \left(-\frac{3}{2}f(-1)-\frac{1}{2}f(0)+\frac{1}{2}f(1)\right)\frac{1}{2}x(x+1) \\ &+2\left(f(-1)-f(1)\right)(1-x^2) \\ &+\left(-\frac{1}{2}f(-1)+\frac{1}{2}f(0)+\frac{3}{2}f(1)\right)\frac{1}{2}x(x-1) \\ &= f(-1)A(x)+f(0)B(x)+f(1)C(x). \end{align} I'll let you write $A,B,C$.

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  • $\begingroup$ so after I multiply whatever I get in front of $f(-1), f(0), f(1)$ on the LHS are $A(x),B(x),C(x)$ right? $\endgroup$ – mandella Feb 10 at 8:34
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    $\begingroup$ @mandella : A is the sum of all terms multiplying f(-1), etc. $\endgroup$ – DisintegratingByParts Feb 10 at 18:39

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