Intermediate Math Problem of the Week 12/27/2017

[PF PotW Robot]

Here is this week's intermediate math problem of the week. We have several members who will check solutions, but we also welcome the community in general to step in. We also encourage finding different methods to the solution. If one has been found, see if there is another way. Occasionally there will be prizes for extraordinary or clever methods. Spoiler tags are optional.

For any continuous real-valued function ##f## defined on the interval ##[0,1]##, let
\begin{gather*}
\mu(f) = \int_0^1 f(x)\,dx, \,
\mathrm{Var}(f) = \int_0^1 (f(x) - \mu(f))^2\,dx, \\
M(f) = \max_{0 \leq x \leq 1} \left| f(x) \right|
\end{gather*}
Show that if ##f## and ##g## are continuous real-valued functions defined on the interval ##[0,1]##, then
$$
\mathrm{Var}(fg) \leq 2 \mathrm{Var}(f) M(g)^2 + 2 \mathrm{Var}(g) M(f)^2
$$

(PotW thanks to our friends at http://www.mathhelpboards.com/)

 

[StoneTemplePython]

I am close on this, with one major loose end. Two questions:

1.) Is it in fact supposed to be a scalar of ##2##, not ##\frac{1}{2}## on the Right Hand Side?

2.) Why did @PotW Tobor change its name?

 

[Delta2]

I am close on this, with one major loose end. Two questions:

1.) Is it in fact supposed to be a scalar of ##2##, not ##\frac{1}{2}## on the Right Hand Side?

2.) Why did @PotW Tobor change its name?

I am unable to comment whether the factor of 2 should be 1/2 but can I ask you something:

Did you prove it first for functions f,g such that ##\mu(f)=\mu(g)=0##?

EDIT: Turns out if f,g are such I can prove it with the factor of 1/2 as you say !

 

[StoneTemplePython]

I am unable to comment whether the factor of 2 should be 1/2 but can I ask you something:

Did you prove it first for functions f,g such that ##\mu(f)=\mu(g)=0##?

EDIT: Turns out if f,g are such I can prove it with the factor of 1/2 as you say !

Exactly. It's straightforward if the random variables are centered (zero mean). It's surprisingly tricky to bootstrap the zero mean case to the general case though.

 

[Delta2]

Ok well, here is my try for the special case ##\mu(f)=\mu(g)=0##
First we notice that ##Var(f)=\mu(f^2)-\mu^2(f)## for any f, so in the special case we consider it will be

##Var(f)=\mu(f^2), Var(g)=\mu(g^2)## and

##Var(fg)=\mu(f^2g^2)-\mu^2(fg)\leq \mu(f^2g^2)## (1)

From (1) we can prove the desired inequality by noticing that for example ##\mu(f^2g^2)\leq \mu(f^2)M(g)^2=Var(f)M(g)^2## so we ll have

##Var(fg)\leq Var(f)M(g)^2## and also ##Var(fg)\leq Var(g)M(f)^2## and by adding these last two we get
##Var(fg)\leq \frac{1}{2} (Var(f)M(g)^2+Var(g)M(f)^2)##.

So we proved a stronger inequality but that's for the special case where the two functions have mean value 0.

 

[PF PotW Robot]

2.) Why did @PotW Tobor change its name?

Just making things a little more clear :)