- #1

Eclair_de_XII

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- 91

- 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)|##.

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)|##.