# Problem Of The Week # 294 - Dec 27, 2017

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#### Ackbach

##### Indicium Physicus
Staff member
Here is this week's POTW:

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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
$\mathrm{Var}(fg) \leq 2 \mathrm{Var}(f) M(g)^2 + 2 \mathrm{Var}(g) M(f)^2.$

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#### Ackbach

##### Indicium Physicus
Staff member
No one answered this week's question, which was Problem B-4 in the 2013 Putnam archive. The solution, attributed to Kiran Kedlaya and associates, follows.

$\newcommand{\Var}{\mathrm{Var}}$

Write $f_0(x) = f(x)-\mu(f)$ and $g_0(x) = g(x)-\mu(g)$, so that $\int_0^1 f_0(x)^2\,dx = \Var(f)$, $\int_0^1 g_0(x)^2\,dx = \Var(g)$, and $\int_0^1 f_0(x)\,dx = \int_0^1 g_0(x)\,dx = 0$. Now since $|g(x)| \leq M(g)$ for all $x$, $0\leq \int_0^1 f_0(x)^2(M(g)^2-g(x)^2)\,dx = \Var(f) M(g)^2-\int_0^1 f_0(x)^2g(x)^2\,dx$, and similarly $0 \leq \Var(g)M(f)^2-\int_0^1 f(x)^2g_0(x)^2\,dx$. Summing gives

\Var(f)M(g)^2+\Var(g)M(f)^2

Now
\begin{align*}
&\int_0^1 (f_0(x)^2g(x)^2+f(x)^2g_0(x)^2)\,dx-\Var(fg) \\&= \int_0^1 (f_0(x)^2g(x)^2+f(x)^2g_0(x)^2-(f(x)g(x)-\int_0^1 f(y)g(y)\,dy)^2)\,dx;
\end{align*}
substituting $f_0(x)+\mu(f)$ for $f(x)$ everywhere and $g_0(x)+\mu(g)$ for $g(x)$ everywhere, and using the fact that $\int_0^1 f_0(x)\,dx = \int_0^1 g_0(x)\,dx = 0$, we can expand and simplify the right hand side of this equation to obtain
\begin{align*}
&\int_0^1 (f_0(x)^2g(x)^2+f(x)^2g_0(x)^2)\,dx-\Var(fg) \\
&= \int_0^1 f_0(x)^2g_0(x)^2\,dx \\
&-2\mu(f)\mu(g)\int_0^1 f_0(x)g_0(x)\,dx +(\int_0^1 f_0(x)g_0(x)\,dx)^2 \\
&\geq -2\mu(f)\mu(g)\int_0^1 f_0(x)g_0(x)\,dx.
\end{align*}
Because of (1), it thus suffices to show that

2\mu(f)\mu(g)\int_0^1 f_0(x)g_0(x)\,dx
Now since $(\mu(g) f_0(x)-\mu(f) g_0(x))^2 \geq 0$ for all $x$, we have