This is a homework question.
Let:
- $V$ : $\mathbb{R}$-vector space of $C^\infty$ functions over the interval $[1,a]$ and are $0$ on the boundary of this interval
- $\langle-,-\rangle$ : be an inner product on V defined as $\langle f,g \rangle = \int^a_1f(x)g(x)e^{x}dx$
- $T \in \operatorname{End}(V)$ : $T(f)=-e^{-x}\frac{d}{dx}(x\frac{df}{dx})$
And I am to show that $T$ is a positive operator. That is, $\langle Tf,f\rangle \ge 0$ $\forall f\in V$. So far, I have tried;
- To find $S \in V$ such that $S^2 = T$. This turned out to be harder than I thought, and I can't think of a systamtic way of doing it
- To directly use the definition of $T$ and the inner product to show $T$ is positive. This boils down to showing $\int^a_1f(f' + xf'')dx \ge 0$
- To show that no solutions to the differential equation $xe^{-x}f''+e^{-x}f'+\lambda f=0$ exist if $\lambda \lt 0$
However, I come up short on all accounts. Perhaps a nudge in the right direction? Thank you.
$\newcommand{\ip}[1]{\left\langle{#1}\right\rangle}$By integration by parts, we have that $$ \ip{f,Tf} = \int_1^a f(x) \left(-e^{-x} \left(xf^\prime(x)\right)^\prime \right) e^{x} dx = -\int_1^a f(x)\left(xf^\prime(x)\right)^\prime dx = -\left.\left(f(x)xf^\prime(x)\right)\right|_1^a+\int_1^a x \left(f^\prime(x)\right)^2 dx. $$ What do you observe about the second term of the final expression? Moreover, given the definition of the vector space $V$, what can you observe about the first term of the final expression?