Equicontinuity in family of functions Let $f, g: $ $\mathbb{R}\to \mathbb{R}$ be continuous functions. For each $y$ in the interval $[0,1]$, consider the function $h_y:$ $\mathbb{R}\to \mathbb{R}$, defined by:
$$
h_y (x):= f (x + g(y))
$$
Is the family of functions equicontinuous and pointwise bounded?
Thank you.
 A: Yes, I think it is pointwise bounded and equicontinuous (but not necessarily uniformly equicontinuous).
Since $g$ is continuous on the compact interval $[0,1]$, it is bounded and we have $M>0$ such that $|g(y)|\leq M$ for all $y\in [0,1]$.
Let us fix  $x_0\in\mathbb{R}$.
Note that $|x_0+g(y)|\leq |x_0|+|g(y)|\leq |x_0|+M=M_0$ for all $y\in [0,1]$, so $z=x_0+g(y)$ belongs to the compact interval $[-M_0,M_0]$. Taking an upper bound $M'$ for $|f|$ on the latter, we get:
$$
|h_y(x_0)|=|f(x_0+g(y))|\leq M' \quad \forall y\in [0,1].
$$
This proves the pointwise boundedness.
Note then that whenever $|x-x_0|\leq 1$, we have 
$$
|x+g(y)|=|x-x_0+x_0+g(y)|\leq |x-x_0|+|x_0|+|g(y)|\leq 1+|x_0|+M=N_0
$$
for all $y$.
So for such an $x$, $z=x+g(y)$ belongs to the compact interval $[-N_0,N_0]$.
Now $f$ is uniformly continuous on the latter.
So for all $\epsilon>0$, there exists $\delta>0$ (make sure to take $\delta \leq 1$ for the argument to work) such that $|z_1-z_2|<\delta$ and $z_1,z_2\in [-N_0,N_0]$ implies $|f(z_1)-f(z_2)|<\epsilon$.
So for $|x-x_0|<\delta$, we have $|(x+g(y))-(x_0+g(y))|=|x-x_0|<\delta$ and $z=x+g(y),z_0=x_0+g(y)\in [-N_0,N_0]$, so
$$
|h_y(x)-h_y(x_0)|=|f(x+g(y))-f(x_0+g(y))|<\epsilon
$$
for all $y\in [0,1]$.
