Adjoint of an operator on an inner product space

65 Views Asked by At

Find the adjoint of $$\left ( Lf \right )\left ( t \right )=t^{2}\frac{df}{dt}+tf$$ where $$f\left ( 0 \right )=1$$ and$$ f'\left ( 1 \right )=2$$ under $$\left \langle f,g \right \rangle=\int_{0}^{1} f\left (t \right )g\left ( t \right )dt$$

Working out I get $$\int_{0}^{1}\left ( t^{2}f'\left ( t \right )g+tf\left ( t \right )g\left ( t \right ) \right )dt$$

The 'second' integral that is the terms $$tf\left ( t \right )g\left ( t \right )$$ is a concern. Can I have some help?

1

There are 1 best solutions below

3
On

The first term is given by integration by parts, the second one by noticing that $$(tf(t))g(t) = f(t)(tg(t)).$$ Therefore, $$\langle tf(t),g(t)\rangle = \langle f(t),tg(t)\rangle$$ so that the operator $f(t)\mapsto tf(t)$.