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I have stuck with following equality, For all $x$, assume function $a(x)$, $b(x)$ have nonzero, and non negative values. (i.e $a(x)>0$, $b(x)>0$, Is the following equality true?

$$\min(ab)=\min(a)\min(b)$$

From the below answers, i need more constraint to achieve, this equality. How about function $a(x)$, $b(x)$ are $C^{n}$ ($n$th differentiable, i.e smooth function) then is this equality holds? (Is the condition for differentiablity is necessary? )

It seems plausible to change this problem into set theory language,

For set $A$ and $B$ with nonzero non-negative elements, assume element $a \in A$, $b \in B$, is this equality true?

$$\min(ab)=\min(a)\min(b)$$

I tried to look up some analysis textbook, but I could not find satisfactory proofs.

phy_math
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    Your set-theoretic reformulation is very different from the original question. For the set-theoretic version, the equality holds (at leasts when sets are finite and nonempty). For the original question -- no, it isn't true. – Dan Shved Jul 27 '14 at 07:58
  • It's probably best to visualize this problem by graphing some functions, rather than casting about for an abstract reformulation. Sometimes that's helpful, but sometimes half a line of understanding is worth a page of algebra. – Joshua P. Swanson Jul 27 '14 at 08:17
  • Your set-theoretic formulation is poorly written. Understood literally, $\min(a)$ means the minimum of $a$... which is $a$. (If you give me just one number, it is the minimum of the numbers you gave me). You probably meant $\min(A)$, which is the minimum of all elements of set $A$. –  Jul 28 '14 at 04:48

4 Answers4

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It's only true if $a(x)$ and $b(x)$ attain their minimum for the same value of $x$.

For instance, consider $a(x) = (x+1)^2+1$ and $b(x) = (x-1)^2+1$.

Clearly, $\min a = \min b = 1$. But $a(x)b(x) = x^4+4$, for which $\min ab = 4$.

EDIT: The OP recently added the following to the question:

From the below answers, i need more constraint to achieve, this equality. How about function $a(x)$, $b(x)$ are $C^{n}$ ($n$th differentiable, i.e smooth function) then is this equality holds? (Is the condition for differentiablity is necessary? )

Adding the constraint that $a(x)$ and $b(x)$ are $C^n$ functions, or even $C^{\infty}$ functions, or even analytic functions won't help, as my above counterexample still meets these additional constraints.

Note that under the given conditions, we can prove that $\min(ab) \ge \min(a)\min(b)$. But as I stated before, equality only holds if there is an $x_0$ such that $a(x_0) = \min(a)$ and $b(x_0) = \min(b)$.

JimmyK4542
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Hint

Just consider $$\Big(a(x)b(x)\Big)'=a'(x)b(x)+a(x)b'(x)$$ If it exists $x_0$ such that $a'(x_0)=b'(x_0)=0$ then,for $x=x_0$, $\Big(a(x)b(x)\Big)'=0$. This is the required condition.

Claude Leibovici
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It is not true. Consider the two functions: $$f(x)=1+|x-1|,\qquad g(x)=1+|x+1|.$$ Obviously $\min f=\min g=1$, but $\min(fg)=3$.

Jack D'Aurizio
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  • Okay, so i think more constraint such as $C^{n}$ is required, then your examples can be eliminated. – phy_math Jul 27 '14 at 13:30
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it's not correct. for example assume $a(x)=x^2 $ and $b(x)=1/x^2$

$min(a)=0$ and $b$ doesn't have minimum but $min(ab)=min(1)=1$

Shabbeh
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