 Solution

For the real lattice version of the Stone-Weierstrass theorem, let $f\in C_{\mathbb{R}}(\Omega )$ and $\epsilon >0$. For any $x,y\in \Omega$, separation of points and constants give $h_{xy}\in L$ matching $f$ at both $x$ and $y$: take an affine combination of a separating function, and for $x=y$ take a constant. Fix $x$. The open sets where $h_{xy}>f-\epsilon$ cover $\Omega$ as $y$ varies. Compactness gives finitely many such functions; their maximum $g_x$ belongs to the lattice $L$, satisfies $g_x>f-\epsilon$ everywhere, and has $g_x(x)=f(x)$. The open sets where $g_x<f+\epsilon$ now cover $\Omega$ as $x$ varies. Choose finitely many and take their minimum $h\in L$. Then $f-\epsilon <h<f+\epsilon$ everywhere. Thus $L$ is uniformly dense in the real continuous functions.

For the complex *-algebra, its real subalgebra $A_{\mathbb{R}}$ of Hermitian C-star elements contains the real and imaginary parts of every member. It separates points, because at least one of those parts distinguishes any pair distinguished by a complex function. Its uniform closure is a unital real subalgebra and is a lattice by the allowed result. The real theorem makes this closure all of $C_{\mathbb{R}}(\Omega )$. Approximating the real and imaginary parts separately proves $A$ is uniformly dense in $C(\Omega )$.

On $[0,1]$, the algebra of polynomials in $t^2$ is unital, closed under conjugation, and separates points because $t\mapsto t^2$ is injective there. Choose $p(t^2)$ within $\epsilon /3$ of $a(t)$, and correct its endpoints byThis is still a polynomial in $t^2$ and has exactly the requested endpoint values. The correction has absolute value at most $\epsilon /3$, because it is a convex combination of two errors of that size. Therefore