Show that $Lip(X)$ is a subalgebra of $C(X)$.

Question

Exercise: Let $X$ be compact. Show that $Lip(X)$ is a subalgebra of $C(X)$.

I'm trying to understand the following solution to this exercise:

Solution:

A linear combination of two Lipschitz functions is again Lipschitz. Indeed, if $f,g:X\to\mathbb{R}$ are Lipschitz and $\alpha,\beta\in\mathbb{R}$, then $h:=\alpha f + \beta g$ satisfies $$\left|h(x) - h(y)\right|\leq \left|\alpha\right|\left|f(x) - f(y)\right| + \left|\beta\right|\left|g(x) - g(y)\right|\leq \left|\alpha\right|K_f d(x,y) + \left|\beta\right|K_g d(x,y) = K d(x,y)$$ with $K = \left|\alpha\right|K_f + \left|\beta\right|K_g$.

The product of two Lipschitz functions is also Lipschitz. To prove let $f,g:X\to\mathbb{R}$ be a Lipschitz function. Let $h = fg$. Recall that $f$ and $g$ are bounded. Now we have the following estimate $$\begin{split}\left|h(x) - h(y)\right| &= \left|f(x)g(x) - f(x)g(y) + f(x)g(y) - f(y)g(y)\right|\\&\leq\left|f(x)\right|\left|g(x) - g(y)\right| + \left|g(y)\right|\left|f(x) - f(y)\right|\\&\leq\|f\|_\infty K_f d(x,y) + \|g\|_\infty K_g d(x,y) = K d(x,y)\end{split}$$ where $K = \|f\|_\infty + \|g\|_\infty$.

Question: Why does this show that $Lip(X)$ is a subalgebra of $C(X)$? I know that a subset $B$ of an algebra $A$ is a subalgebra itself if it is closed under multiplication. If this is being used, I don't understand I have to show that a linear combination of two Lipschitz functions is again Lipschitz.

Thanks!

Answer 1

A subalgebra is a subset that is itself an algebra using the same operations of the bigger algebra. If you recall, an algebra doesn't just have a product: it's also a vector space (and the product must be bilinear). Thus a subalgebra must in particular be a vector subspace, otherwise it wouldn't be an algebra, because it wouldn't be a vector space.

For example consider the subspace $X = \{1\}$ of $C(X)$ which has just the constant function equal to $1$. It's closed under the product, because $1 \cdot 1 = 1$. But it's not a subalgebra, because it's not a vector space: $1+1 = 2 \not\in X$.

Answer 2

"I know that a subset $B$ of an algebra $A$ is a subalgebra itself if it is closed under multiplication."

This statement is false as it stands. This is why you got confused I guess. The right statement is the following:

"A subspace $B$ of an algebra $A$ is a subalgebra itself if it is closed under the bilinear multiplication of $A$."

The intuition behind it is that a vector space together with a closed bilinear multiplication is an algebra. So the corresponding substructure is a subspace with closed bilinear multiplication. I hope it got clearer :)