Questions about $Lip_k[0,1]$

Let $Lip_k[0,1] = \{f \in C([0,1]) : |f(x)-f(y)| \leq k*|x-y|\}$. Prove or disprove the following statements.

1. $Lip_k[0,1]$ is a subalgebra of $C[0,1]$.
2. $Lip_k[0,1]$ is a sublattice of $C[0,1]$.

For part 1, I said that the statement is false. For $f,g \in Lip_k[0,1]$ we have that $|f(x)+g(x)-f(y)-g(y)| \leq 2k*|x-y|$ so it is not a subspace, let alone a subalgebra.

For part 2, my intuition says this is true. But I’m not sure how to proceed. I think it has something to do with $[0,1]$ being compact. Any insight would be appreciated.

The argument you have given for 1) is not valid. If a number is $\leq 2k|x-y|$ there is no guarantee that it is not $\leq k|x-y|$. So you have to prove 1) by giving a counter-example.Let $f(x)=g(x)=kx$ for all x. Then $f,g \in Lip_k$ but $f+g=2f$ is not in $Lip_k$ because $|2f(0)-2f(1)| > k|1-0|$. For 2) consider any $f,g \in Lip_k$. Let $x,y \in [0,1]$ and consider $(f \vee g) (x) -(f \vee g) (y)|$ This number can be $|f(x)-f(y)|$,$|g(x)-g(y)|$,$|f(x)-g(y)|$ or $|g(x)-f(y)|$. In the first two cases we have no problem and the last two cases are similar. So consider the third case where $f(x) \geq g(x)$ and $g(y) \geq f(y)$. We have $f(x)-g(y) \leq f(x)-f(y) \leq k|x-y|$ and $g(y)-f(x) \leq g(y)-g(x) \leq k|x-y|$. Put together, these two inequalities imply that $|f(x)-g(y)|\leq k|x-y|$ as required.