Prove a continuous function multiplied by a positive constant is continuous

[Thas]

I'm having trouble with a certain math problem. Here is the exact problem:


Let f, g: ℝ -> ℝ be continuous functions.
Prove: For all α > 0 and β > 0, the function F: ℝ -> ℝ defined by F(x) = α * f(x) + β * g(x) is continuous.

I am quite certain that I have to solve this using the epsilon delta definition. I have no problems proving a function like "f(x) = 7x² + 5x + 8" is continuous using the epsilon delta definition, but I don't know how to approach this one because the functions f(x) and g(x) aren't formulas like that one.


I understand that the epsilon delta definition means that if the R ↦ R is continuous in x = p, then there exists a δ > 0 for every ε > 0 such that it follows from | x - p | < δ that | f(x) - f(p) | < ε.
It seems to me that in this case, I need to find the δ for which δ > |p - x| implies that ε > |f(p)α-f(x)α|, where it is given that ε > 0 and δ > 0.
And then I get stuck, because I wouldn't know how to find that δ. I can't manage to get any further than |αf(x)-αf(p)|<|αf(p)-αf(p+δ)|.

How should I approach this? Thanks.

 

[daon2]

Let c be a real number in the domains of f and g, let h = af + bg and let e > 0. Set M = max{a,b}

You may find a d1 such that |x-c| < d1 => |f(x)-f(c)| < e/(2M)
You may find a d2 such that |x-c| < d2 => |g(x)-g(c)| < e/(2M)

What can you say about |h(x)-h(c)| now? What delta will work?

 

[pka]

Let f, g: ℝ -> ℝ be continuous functions.
Prove: For all α > 0 and β > 0, the function F: ℝ -> ℝ defined by F(x) = α * f(x) + β * g(x) is continuous.
The epsilon delta definition means that if the R ↦ R is continuous in x = p, then there exists a δ > 0 for every ε > 0 such that it follows from | x - p | < δ that | f(x) - f(p) | < ε.

Because both \(\displaystyle f~\&~g\) are continuous a \(\displaystyle x=c\), consider
\(\displaystyle \begin{gathered}
\left| {\alpha f(x) + \beta g(x) - \alpha f(c) - \beta g(c)} \right| \hfill \\
\leqslant \left| {\alpha f(x) - \alpha f(c) + \beta g(x) - \beta g(c)} \right| \hfill \\
\leqslant \left| {\alpha f(x) - \alpha f(c)} \right| + \left| {\beta g(x) - \beta g(c)} \right| \hfill \\
\leqslant \alpha \left| {f(x) - f(c)} \right| + \beta \left| {g(x) - g(c)} \right| \hfill \\
\end{gathered} \)

By continuity we can ‘control’ both \(\displaystyle \left| {f(x) - f(c)} \right|~\&~\left| {g(x) - g(c)} \right| \)