Prove that the product of a constant and a continuous function is continuous.

Question

My attempt

A function $f$ is continuous if $\forall ε_1>0\,\,\exists δ_1$ such that $|x-a|<δ_1\Rightarrow |f(x)-f(a)|<ε_1$ .

Now we can multiply $ε_1$ with |k| to obtain $|x-a|<δ_1\Rightarrow |k||f(x)-f(a)|<$|k|$ε_1$

$|kf(x)-kf(a)|$ < $|k|ε_1$

Any Positive Real Number can be written as a product of 2 real numbers thus we define $ε_2$=$|k|ε_1$.

Since $\forall ε_1\,\exists δ_1$ and $\forall ε_2\,\exists ε_1$ , we can conclude $\forall ε_2\,\exists δ_1$ where $|x-a|<δ_1\Rightarrow |f(x)-f(a)|<ε_2$

My proof is a bit different from the traditional proof as I use stuff like any positive real number can be represented by product of two positive real numbers instead of letting ε be some value and manipulating it. Also I'm confused whether the last statement should be $|x-a|<δ_1$ or $|x-a|<δ_2$ where $δ_2$ is a new defined $δ$

NOTE- $|k|$ is assumed to be greater than 0 as the proof for k=0 is really easy

Answer 1

This is a very basic proof, but you seem to be confused. Let's take it slowly. Define $g(x)=kf(x)$. This is the sum of a constant and a continuous function alright. Let $\epsilon$ be a positive integer. What we need to show is that there is a $\delta>0$ such that $|x-y|<\delta\Rightarrow |g(x)-g(y)|<\epsilon$. Now, from the continuity of $f$, we deduce that for the positive integer $\epsilon/k$ there is a positive integer $\delta'$ so that $|x-y|<\delta'\Rightarrow |f(x)-f(y)|<\epsilon/k$. Then we need only take $\delta=\delta'$ to have $|g(x)-g(y)|<\epsilon$.

Now, this sort of formal, cumbersome manipulation is really not the point of calculus. As you get acquainted with basic notation, you will soon adopt a more relaxed stance towards this, to the point that you'll be showing that $|g(x)-g(y)|$ is less than some function of $\epsilon$ whose limit at $0$ is $0$ and declairing "well, since $\epsilon$ can be any positive number, that proves the continuity of $g$". You just need to make sure that you practice understanding the image behind the notation.

I also must note that the sequence definition of continuity is a proper way to prove this fact, and you might want to look it up.

Answer 2

Your reasoning is fine but you are mangling the concept that that these proofs must hold for ANY $\epsilon$ and not just a specific one.

To word the proof you want I think you should do this.

$\forall \epsilon > 0$ [It is an abuse of notation to say $\forall \epsilon_1$. THe indexing of $\ _1$ implies you have a specific value in mind but we are making a statement about all possible values..]

$\forall \epsilon > 0, \exists \delta_{\epsilon}$ [The delta is dependent upon the $\epsilon$. Must proofs don't bother with the index for the delta but I like to because it indicates that the $\delta$ value is specific to whatever the $\epsilon$ value is.

So that $|x -a| < \delta_\epsilon \implies |f(x) -f(a)| < \epsilon$.

So $|kf(x)-kf(a)|<|k|\epsilon = \epsilon_2$.

Now your error is in thinking that starting with $\epsilon$ and proving something about a different $\epsilon_2$ is in any way relevant. We must prove something for our first epsilon.

But that is salvageable.

We can say. For any $\epsilon > 0$ there exists an $\epsilon_1$ and a $\delta_{\epsilon_1}$ so that $|k|\epsilon_1 = \epsilon$ and $|x-a|<\delta_{\epsilon_1}\implies |f(x)-f(a)|<\epsilon_1 \implies |kf(x) - kf(a)| < |k|\epsilon_1 = \epsilon$.

==== old ====

For $f$ and for any $\epsilon_1 >0$ you can find a $\delta$ that is dependent upon both $f$ and upon $\epsilon$ (we can call it $\delta_{f, \epsilon_1}$ so that if $|x-a|< \delta_{f, \epsilon_1}$ then $|f(x) -f(a)| < \epsilon_1$.

So for the SAME $\epsilon_1$ we can find an $\epsilon_2$ so that $\epsilon_1 = |k|\epsilon_2$ and $\epsilon_2 = \frac{\epsilon_1}{|k|}$ (assuming $k \ne 0$) and for that $\epsilon_2$ there is a $\delta_{f,\epsilon_2}$ so that if $|x - a| < \delta_{f,\epsilon_2}$ then $|f(x) - f(a)| < \epsilon_2$. And therefore $|k(f(x)) - ka| < |k|\epsilon_2 = \epsilon_1$.

So we have proven for all $\epsilon_1$ (!!!NOT!!! a different $\epsilon_2$ but the FIRST $\epsilon_1$ then there exists a $\delta_{kf, \epsilon_1} = \delta_{f, \epsilon_2}=\delta_{f,\frac {\epsilon_1}{|k|}}$, that does ... the magic bullet stuff.

The thing is you use the new epsilon, to find a new delta, but you use the new delta to apply back to the old epsilon. We can't try to just apply it to a new epsilon altogether.