composition of two uniformly continuous functions.

Solution 1:

• Option $(1):$ True:

  Choose $\epsilon>0.$

  Then $\exists~\delta_1>0$ such that $|x_1-x_2|<\delta_1$$\implies|f(x_1)-f(x_2)|<\epsilon.$

  For above $\delta_1>0~\exists~\delta>0$ such that $|x_1-x_2|<\delta$$\implies|g(x_1)-g(x_2)|<\delta_1.$

  Consequently, $|x_1-x_2|<\delta$$\implies|(f\circ g)(x_1)-(f\circ g)(x_2)|<\epsilon.$

• Options $(2),~(3):$ Not necessarily true:

  Take $f(x)=g(x)=x$ on $\mathbb R.$

  Then $f,g$ are uniformly continuous on $\mathbb R.$

  Note $(f\circ g)(x)=x$ is unbounded and uniformly continuous on $\mathbb R.$

Solution 2:

Your suspicion is justified. The argument you give for the validity of 1 is faulty since it only shows the composite preserves Cauchy sequences, but it does not show that the composite is uniformly continuous. Try working directly with the definition of uniform continuity instead to give a direct proof.

You can make things easier with the counter example to 3: Take $f(x)=g(x)=x$ on $\mathbb R$.