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

• Eclair_de_XII
In summary, for any real number ##\alpha##, if ##f:\mathbb{R}\rightarrow \mathbb{R}## is continuous at ##a\in D\subset \mathbb{R}##, then ##\alpha f## is also continuous at ##a\in D##.
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)|##.

Eclair_de_XII said:
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.

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.

Eclair_de_XII

## 1. What is a scalar multiple?

A scalar multiple is a number that is multiplied by a function. It is a constant value and does not change as the function is evaluated.

## 2. How do you prove that a scalar multiple of a continuous function is continuous?

To prove that a scalar multiple of a continuous function is continuous, we need to show that the function remains continuous at every point in its domain. This can be done by using the definition of continuity, which states that a function is continuous if the limit of the function at a point is equal to the value of the function at that point.

## 3. Can a scalar multiple affect the continuity of a function?

No, a scalar multiple does not affect the continuity of a function. It only scales the values of the function, but does not change its continuity properties.

## 4. Is the product of a continuous function and a scalar multiple also continuous?

Yes, the product of a continuous function and a scalar multiple is also continuous. This is because the product of two continuous functions is also continuous, and a scalar multiple is essentially a constant value that does not affect the continuity of the function.

## 5. What is the importance of proving that a scalar multiple of a continuous function is continuous?

Proving that a scalar multiple of a continuous function is continuous is important because it allows us to manipulate the function without changing its continuity properties. This is useful in many mathematical applications, such as in calculus and differential equations.

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