How to prove that a scalar multiple of a continuous function is continuous

[Eclair_de_XII]

Homework Statement

Suppose that $f:\mathbb{R}\rightarrow \mathbb{R}$ is continuous at $a\in D\subset \mathbb{R}$. Let $\alpha \in \mathbb{R}$. Show that $\alpha f$ is continuous at $a\in D$.

Relevant Equations

If $f$ is continuous at $a\in D$, then for all $\epsilon>0$, there is a $\delta>0$ such that for all $x\in D$ such that $|x-a|<\delta$, $|f(x)-f(a)|<\epsilon$.

Suppose $\alpha=0$. Then $\alpha f=0$, the zero map. Hence, the distance between the images of any two $x_1,x_2 \in D$ through $f$, that is to say, the absolute difference of $(\alpha f)(x_1)=0$ and $(\alpha f)(x_2)=0$, is less than any $\epsilon>0$ regardless of the choice of distance $\delta>0$ between $x_1, x_2\in D$.

Assume now that $\alpha\neq 0$. Let $\epsilon>0$ and set $\epsilon'=\frac{\epsilon}{|\alpha|}$. If $\delta>|x-a|$, then $|f(x)-f(a)|<\epsilon'=\frac{\epsilon}{|\alpha|}$. Equivalently, $\epsilon>|\alpha||f(x)-f(a)|=|\alpha(f(x)-f(a))|=|\alpha f(x) - \alpha f(a)|=|(\alpha f)(x)-(\alpha f)(a)|$.

 

[PeroK]

Homework Statement:: Suppose that $f:\mathbb{R}\rightarrow \mathbb{R}$ is continuous at $a\in D\subset \mathbb{R}$. Let $\alpha \in \mathbb{R}$. Show that $\alpha f$ is continuous at $a\in D$.
Relevant Equations:: If $f$ is continuous at $a\in D$, then for all $\epsilon>0$, there is a $\delta>0$ such that for all $x\in D$ such that $|x-a|<\delta$, $|f(x)-f(a)|<\epsilon$.

Suppose $\alpha=0$. Then $\alpha f=0$, the zero map. Hence, the distance between the images of any two $x_1,x_2 \in D$ through $f$, that is to say, the absolute difference of $(\alpha f)(x_1)=0$ and $(\alpha f)(x_2)=0$, is less than any $\epsilon>0$ regardless of the choice of distance $\delta>0$ between $x_1, x_2\in D$.

Assume now that $\alpha\neq 0$. Let $\epsilon>0$ and set $\epsilon'=\frac{\epsilon}{|\alpha|}$. If $\delta>|x-a|$, then $|f(x)-f(a)|<\epsilon'=\frac{\epsilon}{|\alpha|}$. Equivalently, $\epsilon>|\alpha||f(x)-f(a)|=|\alpha(f(x)-f(a))|=|\alpha f(x) - \alpha f(a)|=|(\alpha f)(x)-(\alpha f)(a)|$.

It's not too bad, but could be better.

 

[member 587159]

No need to consider multiple cases.

Let $\epsilon >0$. Choose $\delta>0$ such that $|f(x)-f(a)| < \epsilon/(1+|\alpha|)$ for $|x-a|<\delta$. Then proceed in the obvious way.