F and G Are Continuous Show That Fg is Co Ntinueous
If f,g are continuous functions, then fg is continuous
Solution 1
Since the topology in ${\bf{R}}$ is still the standard Euclidean, it is still a matter of proving this in the classical way that $|f(x)g(x)-f(x_{0})g(x_{0})|\leq|f(x)-f(x_{0})||g(x)|+|f(x_{0})||g(x)-g(x_{0})|$. Now the boundedness of $|g(x)|$ around $x_{0}$ is given by the continuity of $g$ at $x_{0}$ as well. Note that one could see that $|x-x_{0}|<\delta$ is replaced by some neighbourhood $\mathcal{N}(x_{0})$, and the matter of $\min\{\delta,\delta'\}$ should be replaced by the intersection of two neighbourhoods.
Solution 2
Here is an alternative approach:
Claim 1. If $f: X\to\mathbb{R}$ and $g: X\to\mathbb{R}$ are continuous, then $f+g: X\to\mathbb{R}$ is continuous.
Proof. Given a point $x\in X$ and $\epsilon>0$, we want to show that there exists a neighbourhood $N_{x}$ of $x$ such that $(f+g)(N_{x})\subseteq B_{\epsilon}((f+g)(x))$. By continuity of $f$ and $g$, you can find a neighbourhoods $N'_{x}$ and $N''_{x}$ such that $f(N'_{x})\subseteq B_{\epsilon/2}(f(x))$ and $f(N''_{x})\subseteq B_{\epsilon/2}(g(x))$. Now let $N_{x} := N'_{x} \cap N''_{x}$. Then for each $z\in N_{x}$ we have $$|(f+g)(z)-(f+g)(x)| = |f(z)-f(x)+g(z)-g(x)|\leq |f(z)-f(x)|+|g(z)-g(x)| < \frac{\epsilon}{2}+\frac{\epsilon}{2}=\epsilon$$ so that $(f+g)(z)\in B_{\epsilon}((f+g)(x))$. So $f+g$ is continuous.
Claim 2. If $f:X\to\mathbb{R}$ is continuous, then $f^2: X\to\mathbb{R}$ is continuous.
Here $f^2$ is the function defined by $f^2(x):=(f(x))^2$.
Proof. Given a point $x\in X$, let $\epsilon>0$. Let's assume $\epsilon<1$ for simplicity (this is okay to assume). By continuity of $f$, there exists a neighbourhood $N_{x}$ such that $f(N_{x})\subseteq B_{\epsilon}(f(x))$. Note that for every $z\in N_{x}$, we have $$|f(z)+f(x)|=|f(z)-f(x)+2 f(x)| \leq |f(z)-f(x)|+2|f(x)| \leq \epsilon+2|f(x)| < 1+2|f(x)| = C $$ where $C$ was defined to be that constant $1+2|f(x)|$. Note that $C$ is independent of $z$, and only depends on $x$. Next, for every $z\in N_{x}$, we have
$$|f^2(z)-f^2(x)| = |f(z)-f(x)|\cdot |f(z)+f(x)| \leq \epsilon\cdot C$$ Thus, $f^2(N_{x})\subseteq B_{\epsilon}(f^2(x))$ and $f^2$ is continuous.
Claim 3. If $f:X\to\mathbb{R}$ and $g:X\to\mathbb{R}$ are continuous, then $f\cdot g: X\to\mathbb{R}$ is continuous.
Proof. Notice that $$ f\cdot g = \frac{1}{2}\left[(f+g)^2 -f^2-g^2\right] $$ Now apply claim $1$ to see $f+g$ is continuous. So by Claim 2, all three functions $f^2$, $g^2$ and $(f+g)^2$ are continuous. By Claim $1$ again, $f\cdot g$ is continuous (because it is a sum/difference of these functions).
Solution 3
First, define the function $h:X \to \mathbb{R}^2$ by $h(x)=(f(x),g(x))$; since $f$ and $g$ are continuous then $h$ is continuous.
Now, the multiplication function $\cdot:\mathbb{R}^2 \to \mathbb{R}$, $(a,b) \mapsto a \cdot b$ is continuous. Finally, since $(fg)(x):=(\cdot \circ h):X \to \mathbb{R}$, and the composition of continuous functions is continuous, then $fg$ is continuous.
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Comments
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Let $X$ be a topological space and let $f:X \to R$ ,$g:X \to R$ be continuous functions. Show that $fg$ is continuous.
My work: To show $fg$ is continuous at $x$ for each $x \in X$
Let $y=fg(x)$
To show if $N_y$ is a neighborhood of $y$, then the pre-image of $y$ is a neighborhood of $x$.
I know that there exists $B_\epsilon(y) \in N_y$ so I want to show that $（fg）^{-1}(B_\epsilon(y)) \in N(x)$
Let $N_x=（fg）^{-1}(B_\epsilon(y))$, I want to find an open set in $N_x$.
Can anyone give me a hint of how to choose such open set or idea how to prove this ?
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By showing that $f+g$ is continuous, do we have to show $f-g$ is continuos? Or does that just come for free when showing $f+g$ is continuous?
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@TaylorRendon That is a good question. I believe it follows from the fact that we showed $f+g$ is continuous whenever $f$ and $g$ are continuous. Indeed, if $g$ is continuous, then $-g$ is clearly continuous. Thus, $f+(-g) = f-g$ is also continuous.
Recents
Source: https://9to5science.com/if-f-g-are-continuous-functions-then-fg-is-continuous
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