$f, g:[a, b] \rightarrow [a, b] $ be any two continuous functions $(a, b \in \Bbb{R}) $ such that $\inf f(x) =\inf g(x) $

Solution 1:

Since $[a,b]$ is compact and $f,g$ are continuous, they achieve their infimums over the interval. Thus, we have $\min f = \min g$.

Let $X_f$ and $X_g$ be the set of minimums for $f$ and $g$ respectively. That is,

$$X_f := \{ x\in [a,b]: \forall y \in [a,b], \ f(x) \le f(y) \}.$$

If $X_f \cap X_g \neq \emptyset$. Then we have a point where they are equal. Otherwise let us look at the function $h(x) = f(x) - g(x)$. Since the minimums do not coincide, we have that $h(x)$ is negative on $X_f$ and is positive on $X_g$. Further, we have that $h$ is continuous.

Then using the intermediate value theorem, say that $h(x_0) = 0$ for some $x_0 \in [a,b]$. Thus, we have proved the existence of a point $x_0$ where $f(x_0) = g(x_0)$.

Your example then disproves the other 3 statements.