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So basically, the question is pretty straight forward. But I’m having troubles proving or countering the following statement, because of the constant k.

If $f:D\rightarrow \Bbb R$ is continuous on a topological space $D$ and $k\in \Bbb R$, then $kf$ is continuous on $D$.

Seems to me that I have to use the theorem for multiplication of two products. But $k$ is a constant. So how do I exactly apporach this matter? I know for a fact that any function multiplied by a constant is still continuous if and only if the function itself is continuous.

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Ariyan Jahanyar
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    I’m sorry! It should have been f, but sometimes my phone uses capitals.. – Ariyan Jahanyar Dec 06 '18 at 18:56
  • Do you know how to prove that $g(x)=kx$ is continuous? Then your function is a composition of $g$ with $f$. If you want an explicit proof then there are several alternatives depending on what $D$ is and what you mean by "continuous". – freakish Dec 06 '18 at 19:00
  • It is not given what D is, the question is given like that. But am I allowed to say that g(x)=kx? Since I don’t know how g(x) is defined. – Ariyan Jahanyar Dec 06 '18 at 19:03
  • I just defined $g$ by $g(x)=kx$ for you. -.- It's a simple linear function, dude. – freakish Dec 06 '18 at 19:34

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Let $f:D\to\mathbb{R}$ be a continuous function. Now, you have that $g_\lambda:\mathbb{R}\to \mathbb{R}:x\mapsto \lambda x$ is continuous for every $\lambda\in\mathbb{R}$. Then, consider $h=g_\lambda\circ f:D\to\mathbb{R}$ that is continuous since is the composition of two continuous functions, and $h(x)=\lambda x$.

José Alejandro Aburto Araneda
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