Showing thislinear operator on an inner product space is its own transpose

Question

Let $H$ be the inner product space of continuous real valued functions defined on $[0,1]$ where $(\alpha\mid\beta)=\int_{0}^{1} \alpha(u)\beta(u)du$

Put $K(s,t)=\min\{s,t\}-st$. Define $T∈L(V,V)$ by

$\int_{0}^{1} K(s,t) \alpha(t) dt$

How do I show that $T^t=T$? I need to show that $(T(\alpha)\mid\beta)=(\alpha\mid T(\beta))$ for all $(\alpha,\beta)∈H^2$, correct?

Answer 1

I need to show that $(T(\alpha)\mid\beta)=(\alpha\mid T(\beta))$ for all $(\alpha,\beta)∈H^2$, correct?

Yes. To see this, write \begin{align} (T(\alpha)\mid\beta)&=\int_0^1T(\alpha)(s)\beta(s)\mathrm ds\\ &=\int_0^1\left(\int_0^1 \alpha(t)\min\{s,t\}\mathrm dt\right)\beta(s)\mathrm ds\\ &=\iint_{[0,1]^2} \alpha(t)\beta(s)\min\{s,t\}\mathrm dt\mathrm ds. \end{align}
Then change the order of integration.

Note that it would work more generally if $K(s,t)=K(t,s)$ for each $s$ and $t$.