Let $V$ be the vector space of all polynomials with degree less than or equal to $3$. Let $V$ have the inner product, $$\langle f|g\rangle=\int_0^1f(t)g(t)\,dt.$$ Let $D$ be the differentiation operator on $V$. Find $D^{*}$.
(Note that $V$ is finite-dimensional.)
I'm confused. I can get a matrix for $D$ in the standard basis and take the conjugate transpose of that to find $D^{*}$. But then I'm not using the inner product, and am restricting myself to work in a fixed basis. I'm not sure what exactly the question is asking me to do. Any help would be appreciated.
This may not be the method you are thinking of, but perhaps this shows you how one typically uses the inner product to find adjoints.
We define the adjoint operator $D^*$ of $D$ as the operator satisfying \begin{equation} \langle Df, g \rangle = \langle f, D^* g\rangle \end{equation} for all $f,g \in V$.
So, notice that given any $f,g\in V$ we have \begin{align} \langle Df,g\rangle &= \int_0^1 \frac{d}{dx} f(t) g(t) dt\\ &= -\int_0^1 f(t) \frac{d}{dx}g(t) dt + f(1)g(1)-f(0)g(0). \end{align} We can write the boundary terms using delta functions as $$ f(1)g(1) = \int_0^1 \delta(t-1)f(t)g(t)dt $$ and $$ f(0)g(0) = \int_0^1 \delta(t)f(t)g(t)dt $$ so that overall we have $$ \langle Df,g\rangle = \int_0^1 f(t)\left(-\frac{d}{dx} +\delta(t-1)-\delta(t)\right)g(t)dt. $$ Thus, the adjoint operator is $$ D^* = -\frac{d}{dx} + \delta(t-1)-\delta(t). $$