Let $f_1, f_2$ be two functions from $\mathbb R\to\mathbb R$. Suppose that $f_1$ and $f_2$ are both continuous at $x_o∈ \mathbb R$. Let $g(x)=\min(f_1(x),f_2(x))$. Prove $g(x)$ is continuous at $x_o$. Has to be proven using the epsilon-delta definition of continuity: $∀ ε>0 ∃ δ>0$ such that if $x∈I$ and $|x−x_o|<δ$ then $|f(x)−f(x_o)|<ε$.
Asked
Active
Viewed 885 times
2 Answers
4
Not sure who down voted because I think its a fair question.
Since both $f_1$ and $f_2$ are continuous at $x_0$, we have the following relations. $\forall\epsilon_1>0$, there exists a $\delta_1$ such that $|x-x_0|<\delta_1\implies |f_{1}(x)- f_{1}(x_{0})|<\epsilon_1$ (Eq1).
Similarly we have that for $\forall\epsilon_2>0$, there exists a $\delta_2$ such that $|x-x_0|<\delta_2\implies |f_{2}(x)-f_{2}(x_0)|<\epsilon_2$ (Eq2).
Let $g(x)=min(f_{1}(x),f_{2}(x))$. Without loss of generality suppose that $f_{1}(x_0)<f_{2}(x_{0})$.
Then at $x_0$ we have that $g(x_0) = f_{1}(x_{0})$. To show that $g(x)$ is continuous we need to show that $\forall\epsilon>0$ there exists $\delta$ such that $|x-x_0|<\delta\implies |g(x)-g(x_0)|<\epsilon$.
Now as we have assumed $f_{1}(x_0)<f_2(x_0)$ we need $|g(x)-f_{1}(x_0)|<\epsilon|$.
If $g(x)$ is always equal to $f_1(x)$ (that is $f_1<f_2$ for all $x$, then we are done, since $f_1$ is continuous). If not, note that $|f_2(x)-f_1(x_0)|< |f_2(x)-f_2(x_0)|<\epsilon_2$ by (Eq2). This tells us how to choose a $\delta$. That is for each $\epsilon >0$ choose $\delta = min\{\epsilon_1,\epsilon_2\}$.
Don Shanil
- 363
0
Let $x_n \rightarrow x_0$. Suppose w.l.o.g. that $g(x_0) = f_1(x_0)$.
Since $f_1$ is continuous we know that $f_1(x_n) \rightarrow f_1(x_0)$, and $g(x_n) \le f_1(x_n)$, hence we can conclude that $\limsup g(x_n) \le g(x_0)$.
Also by continuity, $f_2(x_n) \rightarrow f_2(x_0) \ge g(x_0)$. Hence we can conclude that $\liminf g(x_n) \ge g(x_0)$.
Combining these we get $g(x_n) \rightarrow g(x_0)$.
Deven Ware
- 7,076