Show that the operator is linear

Question

Consider the vector space of polynomials with degree less than or equal to $3$, $P_3$, equipped with inner product, $$\langle f|g\rangle=\int_0^1f(t)g(t)\,dt.$$

I am asked to find the adjoint of the differentiation operator, $D=\frac{d}{dt}$. I find that $$D^*=\delta(t-1)-\delta(t)-D,$$ where $\delta$ refers to the Dirac delta function. However, I need to verify that this is indeed a linear operator on $P_3$, which I'm having trouble doing. Namely, I need to show that the delta function is a linear operator on $P_3$.

Answer 1

You need to show that

$$ D^*(\alpha f + \beta g) = \alpha D^* f + \beta D^* g $$

To do that note that

• $\delta(t) [\alpha f(t) + \beta g(t)] = \alpha \delta(t) f(t) + \beta \delta(t) g(t)$
• $D$ is linear

so that

\begin{eqnarray} D^*(\alpha f + \beta g) &=& \left(\delta (t-1) - \delta(t) -D \right) (\alpha f + \beta g) \\ &=& \left(\delta (t-1) - \delta(t) -D \right) (\alpha f) + \left(\delta (t-1) - \delta(t) -D \right)(\beta g) \\ &=& \alpha \left(\delta (t-1) - \delta(t) -D \right) f + \beta (\left(\delta (t-1) - \delta(t) -D \right))g \\ &=& \alpha D^* f + \beta D^* g \end{eqnarray}