Upper bound for integral of two variables

Question

Is it true that if $f(x,y)$ and $g(x,y)$ are two positive real bounded functions in $\mathbb{R}^2$, then $$ \int \int dx \, dy \, f(x,y) \, g(x,y) \leq \sup_{ \tilde y \in \mathbb{R}} \, \, \int \int dx\, dy \, f(x,y) \, g(x,\tilde y) \, \, $$ assuming that the integral in the left hand side is finite? I am confused.

Answer 1

Big idea: these integrals, at least over a small domain, and for simple $f$ and $g$, are analogous to double-sums, which are easier to think about. If we look at $$ \sum_{i=1}^n \sum_{j=1}^n f_{ij} g_{ij} \le \sup_k \sum_{i=1}^n \sum_{j=1}^n f_{ij} g_{ik}, $$ where $f$ and $g$ are now matrices, we can divide both sides by $n$ to get $$ \frac{1}{n}\sum_{i=1}^n \sum_{j=1}^n f_{ij} g_{ij} \le \sup_k \frac{1}{n}\sum_{i=1}^n \sum_{j=1}^n f_{ij} g_{ik}, $$

We can now think about each matrix as made up of $n$ column vectors, $f_k$ and $g_k$. The left hand side (which I'll call $L$) then is the average of the pairwise dot-products: $$ L = \frac{1}{n} \sum_j f_j \cdot g_j $$ The right hand side (denoted $R$), on the other hand, can be reshuffled a little to become \begin{align} R &= \sup_k \frac{1}{n}\sum_{i=1}^n \sum_{j=1}^n f_{ij} g_{ik}\\ &= \sup_k \frac{1}{n}\sum_{i=1}^n \bigl( \sum_{j=1}^n f_{ij} \bigr) g_{ik} \\ &= \sup_k \sum_{i=1}^n \bigl( \frac{1}{n}\sum_{j=1}^n f_{ij} \bigr) g_{ik}\\ &= \sup_k \bar{f} \cdot g_k, \end{align} which is the biggest dot product between the average of the $f$ vectors, $\bar{f}$ and any column of $g$.

So the question becomes: is the average of $f_j \cdot g_j$ always less than the max (over $k$) of $\text{avg}(f_j) \cdot g_k$? And the answer to that is "no". We can see this in the $2 \times 2$ case. Let $$ F = G = \pmatrix{0 & 2 \\ 2 & 0} $$

(more generally, we can let $F$ be an $n \times n$ matrix of all $n$s, with $0$s on the diagonal, but I'll stick with $2 \times 2$ here). The average column of $F$ is then $\bar{f} = \pmatrix{1\\1}$. $L$ is $\frac{4 + 4}{2} = 4$. $R$ is ... well, $\bar{f} \cdot g_k$ is independent of $k$: it's just $2$. So $R = 2$. And we end up with $L > R$.

Now we have to turn this intuition into a function.

Restrict the domain to $D = [0, 2] \times [0, 2]$ by defining $f$ and $g$ to be zero outside $D$. Now we can write everything with finite integrals. I know; this makes $f$ and $g$ nonnegative rather than positive. I'll come back to that at the end.

Next, I'm going to make $f$ and $g$ piecewise constant on little squares (indeed, 1 x 1 squares). So now we can describe $f$ and $g$ via $2 \times 2$ matrices indicating the value in each little square. The value in the lower-left square $ 0 \le x, y \le 1$, for instance, will be the lower left entry of the matrix $F$ above, i.e., $2$; the value in the upper left square $0 \le x \le 1, 1 \le y \le 2$, will be the upper left entry of $F$, namely $0$, and so on.

Yes, this function is non-negative rather than positive. You may replace the $0$ in this case with $.000001$, and then work out the remaining details. It's not going to turn $2$ into something greater than $4$.

Now the integrals you need to compute are just $4$ times the sums I computed above, hence we end up with the left integral being larger than the right integral, and we're done.

To more properly extend $f$ and $g$ outside the square, define them via $$ f(x, y) = \begin{cases} \text{as above} & 0 \le x, y \le 2 \\ \frac{C}{(x-1)^{100} (y-1)^{100}} & \text{otherwise} \end{cases} $$ By picking $C$ small, you can arrange for the integral of $f^2$ over the exterior of $D$ to be no larger than, say, $0.0001$, and the whole argument still holds.