Consider a Fredholm operator $K$ define by $$Ku(x)=\int_{a}^{b} k(x,y)u(y) \ dy \ \ \ a\leq x\leq b,$$ where $k(x,y)$ is some known function of $x$ and $y$ called the kernel of $K$, and define the usual inner product $$\langle f,g \rangle=\int_{a}^{b} f(x)g(x) \ dx.$$ Te adjoint of $K$ is the integral operator $K^*$ define by $$K^*u(x)=\int_{a}^{b} k(y,x)u(y) \ dy \ \ \ \ a\leq x\leq b.$$ Show that $\langle Ku,v \rangle=\langle u,K^{*}v \rangle$ for any continuous function $u,v$.
My attempt:
Consider \begin{align} \langle Ku,v \rangle&=\int_{a}^{b} Ku(x)v(x) \ dx \\ &=\int_{a}^{b} \int_{a}^{b} k(x,y)u(y) v(x) \ dy \ dx \\ \end{align} But I am not certain on how to proceed, as I assume $k(x,y)\neq k(y,x)$. A hint would be very helpful.