# How to prove that $L_{D}\leq L_{U}$?

I have the following two languages: $$L_{U}\triangleq\left\{ \langle M,x\rangle\,:\,M\text{ accepts }x\right\} ,L_{D}\triangleq\left\{ \langle M\rangle\,:\,M\text{ accepts }\langle M\rangle\right\}$$ I'm trying to prove that $$L_{D}\leq L_{U}$$. For that I need to show that there exists $$f\,:\,\Sigma^{*}\to\Sigma^{*}$$ total, can be calculated by a Turing machine and follows: $$\forall x\in\Sigma^*$$, $$x\in L_D$$ iff $$f(x)\in L_U$$.

Let there be $$f\,:\,\Sigma^{*}\to\Sigma^{*}$$ so $$f\left(\langle M\rangle\right)=\langle M,M\rangle$$. I have proved that $$f$$ is total and also follows the last condition. All that left to do is to prove that $$f$$ can be calculated by a Turing machine. How should I describe the steps of machine $$M_f$$ on $$\langle M\rangle$$?