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Given two elements $f,g$ from the vector space $\mathbb{R}[x]$, we define the inner product to be $$\langle f,g \rangle = \int ^1 _0 fg \,\,dx.$$ If $Df$ is the derivative of $f$, prove that $D$ doesn't have an adjoint, i.e., there does not exist a $D^*$ such that $$\langle Df,g \rangle = \langle f, D^*g \rangle.$$

I'm not sure exactly how to begin. Does the bound $[0,1]$ play a part in the adjoint not existing? The only time I came across such an inner product is in the space of $L^2$ functions where we can define $$\langle f,g \rangle = \int f \bar{g}.$$

But even so, this is a linear algebra exercise so I don't think we can assume $f,g$ to be in $L^2(\mathbb{R})$. Any help with some direction would be appreciated.

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  • $\begingroup$ The definition of the inner product involves only the values on $[0,1]$ and $\langle .,. \rangle$ is not even an inner product if the domain is taken as $\mathbb R$. So it is necessary to restrict to the unit interval in which case your functions belong to $L^{2}[0,1]$. $\endgroup$ Apr 4, 2022 at 5:04
  • $\begingroup$ @KaviRamaMurthy I hope clarify something: in my notes, it says that we can define the inner product inside $L^2(E)$ for $E$ a positive measure subset of $\mathbb{R}^n$ by $$\langle f,g \rangle = \int f\bar{g}.$$ I think it should be written as bounded positive measure instead, otherwise the integral might be infinite? $\endgroup$
    – oleout
    Apr 4, 2022 at 5:18
  • $\begingroup$ The question involves only integrals w.r.t Lebesgue measure on $(0,1)$. $\endgroup$ Apr 4, 2022 at 5:21
  • $\begingroup$ Yes but I'm wondering if it can still be asked if we change to, say, $[0,2]$. I'm trying to figure out if the bound here could explain why the adjoint of derivative doesn't exist. I've no experience with this type of question. $\endgroup$
    – oleout
    Apr 4, 2022 at 5:24
  • $\begingroup$ The adjoint doesn't exist for any compact interval $[a,b]$. $\endgroup$ Apr 4, 2022 at 5:29

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Let's assume by contradiction that $D$ has an adjoint. In particular, if we set $u = D^{*}(1)$ then we must have the identity $$ f(1) - f(0) = \int_0^1 f'(x) \, dx = \left< Df, 1 \right> = \left< f, D^{*}(1) \right> = \left< f, u \right> $$ for all $f \in \mathbb{R}[x]$. Taking absolute value and using the Cauchy-Schwarz inequality, we get that $$ |f(1) - f(0)| \leq \| f \| \cdot \| u \| = \left( \int_0^1 f^2(x) \, dx \right)^{\frac{1}{2}} \|u\|. $$ Now by plugging $f(x) = x^n$ for $n > 0$, we get that $$ 1 \leq \frac{\|u\|}{\sqrt{2n+1}} $$ for all $n > 0$. This implies that $\sqrt{2n+1} \leq \| u \|$ for all $n > 0$ which is absurd.

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    $\begingroup$ The idea behind the argument is this: Fix some $g$ and consider the linear functional $f \mapsto \left< Df, g \right>$. In order for $D$ to have an adjoint, this linear functional should be represented by some vector (which we then call $D^{*}g$) and this should hold for all $g$. To show that $D$ doesn't have an adjoint, it is enough to choose one such $g$ and show that $\left< Df, g \right>$ is not representable so I chose $g = 1$. A necessary condition for the linear functional to be represented is for it to be continuous but $f \mapsto \left< Df, 1 \right> = f(1) - f(0)$ is not $\endgroup$
    – dejavu
    Apr 4, 2022 at 14:32
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    $\begingroup$ continuous (with respect to the norm induced by the integral inner product) which is what my contradiction shows. $\endgroup$
    – dejavu
    Apr 4, 2022 at 14:33
  • $\begingroup$ Nice choices of $g$ and $f$ to make this answer easy to read and comprehend, thank you. $\endgroup$
    – oleout
    Apr 4, 2022 at 15:28

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