If the product of two continuous functions is zero, must one of the functions be zero?

Question

Suppose that I have two continuous functions

$$f : \left[ a, b \right] \rightarrow \mathbb{R} \quad \text{and} \quad g : \left[ a, b \right] \rightarrow \mathbb{R}$$

and they have the following property

$$f(x) \times g(x) = 0 \space , \forall x \space \in \left[ a, b \right]$$

Can I say that one of the functions necessarily has to be equal to $0$?

For example, $f(x) = 0 \space , \forall x \space \in \left[ a, b \right]$.

UPDATE: Ok, I can see from the counterexamples that the affirmation is not true, but now I cannot see in which cases it is true. If I let the function $g : \left[ a, b \right] \rightarrow \mathbb{R}$ be any continuous function, then in that case must I have $f(x) = 0$ ?

Answer 1

This is not true in general. Consider \begin{equation} f(x) = \left\{\begin{array}{cc} 0 & a \leq x \leq \frac{a+b}{2} \\ x - \frac{a+b}{2} & \frac{a+b}{2} < x \leq b\end{array}\right. \end{equation} and \begin{equation} g(x) = \left\{\begin{array}{cc} x - \frac{a+b}{2} & a \leq x \leq \frac{a+b}{2}\\ 0 & \frac{a+b}{2} < x \leq b \\ \end{array}\right. \end{equation} Both are clearly continuous and not identically zero on $[a, b]$, yet $f(x)g(x) = 0$ for every $x \in [a, b]$.

Edit: In response to the updated question, the answer is yes. If we want $f(x)g(x) = 0$ for all $x \in [a, b]$ and for any arbitrary continuous function $g$, then it must be true that $f \equiv 0$ on $[a, b]$.

Answer 2

No. Imagine a little "bump" around $0$, and a different "bump" around $10$. If you want something more precise, I mean a bump. Then their product will always be $0$ even though neither function is $0$.

Answer 3

That cannot be concluded. For example if $[a,b]=[-1,1]$ let $f(x)=x+|x|$ and $g(x)=x-|x|.$ Note if $x \ge 0$ then $g(x)=0$ while if $x \le 0$ then $f(x)=0$ so the product is zero at any $x$ in $[-1,1],$ yet neither function is the zero function.

Answer 4

Consider that the reals are an integral domain, so there exist no zero divisors.

Let $x\in[a,b]$, where $a,b\in\mathbb R$.
Consider a point in $f(x)g(x)$.
Assume $f(x)\ne0$.
Now $f(x)g(x)=0 \Rightarrow g(x)=0$, since $\mathbb R$ has no zero divisors.
Repeat for the assumption that $g(x)\ne0$.

So $f(x)g(x)=0 \Rightarrow f(x)=0 \lor g(x)=0$.

All that is required for the product to be zero throughout the target set is that either image is zero $\forall x\in[a,b]$.

Answer 5

They do not both have to always be zero, but one has to be zero whenever the other is not. Suppose that $\exists x$ s.t. $f(x) \ne 0$ and $g(x) \ne 0$ then clearly $f(x)g(x) \ne 0$.

Answer 6

No. For instance, take the functions, in $[-1,1]$ defined as $$f(x)=\begin{cases}0 && \text{if } x<0\\x &&\text{if }x\geq 0\end{cases}$$ $$g(x)=\begin{cases}x && \text{if } x<0\\0 &&\text{if }x\geq 0\end{cases}$$ which have zero product everywhere, but are neither constant. It is of note that, at $x=0$, both are $0$.

You can say that, if $S_f$ is the set of $x\in[a,b]$ such that $f(x)=0$ and $S_g$ is defined analogously, then $S_f\cup S_g=[a,b]$ and, moreover, if neither is empty, then $S_f\cap S_g$ is not empty either. (This follows from the fact that both sets are closed, and you cannot partition a closed interval into more than one disjoint closed set - they must intersect, at least, on their boundary)

Answer 7

you can say that ${\mu(f=0)}\geq\frac{b-a}{2}$ or ${\mu(g=0)}\geq\frac{b-a}{2}$ where $\mu$ denotes the lebesgue measure.

Answer 8

Just think about this. For each value of x, the images of the functions are both reals. Yes, just real numbers being multiplied! That can't be zero at least one of them is zero. So the functions has to obey that rule in [a,b].

Edit: sorry, I misinterpreted the question.