# Comparing two functions with same trend

Let $$f$$ and $$g$$ are two given functions with same trend. What I mean by same trend is following:

$$\forall x_1, x_2 \quad f(x_1) \leq f(x_2) \implies g(x_1) \leq g(x_2)$$ essentially, the above equation says that if $$x_1$$ and $$x_2$$ are two points such that $$f(x_2)$$ is greater than $$f(x_1)$$ then it is also the case that $$g(x_2)$$ is greater than $$g(x_1)$$.

If I know the maximum ($$x_* | \forall x f(x) \leq f(x_*)$$) for one of the functions (say $$f$$), is it always possible to bound the other function ($$g$$ in our case) by some simple transformation $$T$$ of $$f(x_*)$$ i.e.

$$if \quad f(x_*) = \max_x f(x)$$ $$is \quad \max_x g(x) \leq T(f(x_*)) \quad?$$ where $$T$$ is some transformation like translation.

In one dimension, I believe we can always translate $$f$$ to be always above $$g$$. To give more context, this comes in a multi fidelity setting where one of the function is an approximation of other where the two function values might not be of same scale but follow a "same trend" as defined above.

I don't know if I understood the question correctly, but it looks trivial to me. Let $$f=0$$ and $$g$$ be a constant. Then the hypothesis is satisfied. If $$T(f{(x_{*})})$$ any thing that depends only on $$f$$ but not on $$g$$ then we cannot expect $$\max g(x) \leq T(F(x_{*}))$$ since $$g$$ is an arbitrary constant.
• Thank you for the answer. Yes, $T(f(x_*))$ can only depend on $f$ and not on $g$. Sep 2, 2019 at 5:50