How to subtract infimum of a set from infimum of another set?

Question

Suppose $A \subset M$ where $M$ is a metric space. Let $f:A \rightarrow \mathbb{R}$ be a Lipschitz map and $g:M \rightarrow \mathbb{R}$ to a map.

Define $$g(y) = \inf_{x \in A}{\{ f(x) + \| f \|_{Lip} d(x,y) \}}$$

Is the last equality correct?

$$|g(y) -g(y^\prime)| $$ $$=| \inf_{x \in A}{\{ f(x) + \| f \|_{Lip} d(x,y) \}} -\inf_{x \in A}{\{ f(x) + \| f \|_{Lip} d(x,y^{\prime}) \}}|$$ $$= \inf_{x \in A}\{ \| f \|_{Lip} d(x,y) - \| f \|_{Lip} d(x,y^{\prime}) \} $$

UPDATE: Okay, so the equality above is not true. But since $\inf(A) - \inf(B) \leq \sup(A-B)$, can we say the following?

$$=| \inf_{x \in A}{\{ f(x) + \| f \|_{Lip} d(x,y) \}} -\inf_{x \in A}{\{ f(x) + \| f \|_{Lip} d(x,y^{\prime}) \}}|$$ $$ \leq \sup_{x \in A}\{ \| f \|_{Lip} d(x,y) - \| f \|_{Lip} d(x,y^{\prime}) \} $$

Actually my aim is to show that $\| g \|_{Lip} = \| f \|_{Lip}$. I think I need the following equality:

$$\inf(A-B) \leq \inf(A) - \inf(B) \leq \sup(A-B).$$

Answer 1

No. In general, we don't have

$$ \inf (A-B) = \inf A - \inf B $$ In fact, $\inf(A-B) = \inf(A)-\sup(B)$.

See here for a similar question, which gives a link to these lecture notes.

I would suggest possibly taking a convergent sequence approach to your problem, i.e. let $(a_n)\rightarrow g(y)$ and $(a_n^\prime)\rightarrow g(y^\prime)$. Then study the sequence $(a_n-a_n^\prime)$. No guarantee this will work, but it feels right.