Sum of uniformly continuous fuctions

Question

If the statment is false (give a counter-exemple), else if the statment is true (Prove it);

Let $f:\mathbb{R} \to \mathbb{R}$ and $g:\mathbb{R}\to \mathbb{R}$. If $f$ and $g$ are both uniformly continuous and $\lambda \in \mathbb{R}$

$(1)$ $h(x) = f(x) + \lambda g(x)$ is uniformly continuous for all $x \in \mathbb{R}$.

For $(1)$ i imagine that (for a supoused $\epsilon_{f+g} > 0$) somehow $\delta_f$ sum with $\lambda \delta_g$ does´t change the uniformity of the function... well i don´t know if this idea make sense and i'm having troubles formalizing it (with $\epsilon$ and $\delta$). Maybe someone has a clearer and more formalized idea that can help me.

Answer 1

If $\lambda=0$, then the statement is trivial. I will assume that $\lambda\neq0$.

Take $\varepsilon>0$. There is a $\delta_1>0$ such that$$\lvert y-x\rvert<\delta_1\implies\bigl\lvert f(y)-f(x)\bigr\rvert<\frac\varepsilon2$$and there is a $\delta_2>0$ such that$$\lvert y-x\rvert<\delta_1\implies\bigl\lvert g(y)-g(x)\bigr\rvert<\frac\varepsilon{2\lvert\lambda\rvert}.$$So, if $\delta=\min\{\delta_1,\delta_2\}$,\begin{align}\lvert y-x\rvert<\delta&\implies\lvert y-x\rvert<\delta_1\text{ and }\lvert y-x\rvert<\delta_2\\&\implies\bigl\lvert\bigl(f(y)+\lambda g(y)\bigr)-\bigl(f(x)+\lambda g(x)\bigr)\bigr\rvert<\varepsilon.\end{align}

Answer 2

Let $\varepsilon > 0$. Then, there are $\delta_1, \delta_2 > 0$ such that. For any $x, y \in R$ with $|x - y| < \delta_1$ we have $|f(x) - f(y)| < \varepsilon$ and if $x, y \in R$ is such that $|x - y| < \delta_2 $ we have $|g(x) - g(y)| < \varepsilon$. Choosing $ \delta = \min\{\delta_1, \delta_2\} > 0$, we have that if $x, y \in R$ and $|x - y| < \delta $ then: $$|h(x) - h(y)| \leq |f(x) -f(y)| + \lambda |g(x) - g(y)| \leq (1 + \lambda) \varepsilon.$$ So, $h$ is uniformly continuous too.

Moreover: It is not true that $f, g$ uniformly continuous implies $f/g$ uniformly continuous: If $X = (0, \infty)$ then $f(x) = x$ and $g(x) = |x|$ are uniformly continuous on $X$, but $h = f/g$ is not.