During a quest to find an appropriate solution to a problem, I have been reduced to the following exercise:

Assume two compactly-defined and Lipschitz-continuous functions $A_1(x)$ and $A_2(x)$ in the domain $x \in \Xi \equiv [0,0.2]$ such that:

$ A_1(x) < 0 \quad \forall x \in \Xi $

$ A_2(x) < 0 \quad \forall x \in \Xi $


$ \int_\Xi (A_1+A_2)dx = -\frac{8\sqrt{2}}{10} $

$ \int_\Xi (A_1^2+A_2^2)dx = \frac{16}{5} $

Prove that $A_1(x) = A_2(x) = \frac{-4}{\sqrt{2}} \quad \forall x \in \Xi$ is a unique solution satisfying all above conditions.

So far, I have approached the problem as a geometrical problem in infinite dimension. For example, find $\boldsymbol{A}$ such that:

$\boldsymbol{A} \in \mathbb{R}_{< 0}^N \quad N \in \mathbb{N} $

$ \sum_{i=1}^N A_i = -4/\sqrt{2}N $

$ || \boldsymbol{A} ||^2 = 8N $

trying to show that $A_i = -4/\sqrt{2} \quad \forall i = 1,2,...,N$ is the only solution for specific integer $N$s. However, this becomes a difficult problem for me in $ N \geq 4 $.

Also, some numerical experimentation using optimization algorithms seems to validate the above conclusion. However, the problem is that I cannot really check the space of all numbers using numerical methods.


This is a basic calculus problem. No need to "approach the problem as a geometrical problem in infinite dimension", whatever that means.

We rename $\Xi$ to $Z$ for convenience. Consider the function $F = (A_1+2\sqrt{2})^2+(A_2+2\sqrt{2})^2$. We calculate it's integral:

$$ \int_Z (A_1+2\sqrt{2})^2 + (A_2+2\sqrt{2})^2 = \int_Z A_1^2 + 4\sqrt{2}A_2 + 8 + A_2^2 + 4\sqrt{2}A_2 + 8 $$ $$ = \int_Z (A_1^2+A_2^2) + 4\sqrt{2}(A_1+A_2) + 16 = \frac{16}{5}+ 4\sqrt{2}(\frac{-4\sqrt{2}}{5})+\frac{16}{5}$$ $$ = \frac{32}{5}-\frac{32}{5} = 0$$

This implies that $F$ is $0$ on $Z$ except possibly on a set of measure $0$. Since Lipschitz continuous on a compact domain in $\Bbb R$ implies continuous, $F$ is continuous and $F$ is actually identically $0$ on $Z$. Since it is a sum of two nonnegative functions, this implies that each of those functions are identically zero on $Z$, or that $A_1=A_2=-2\sqrt{2}$ on $Z$.

  • $\begingroup$ How would you approach the problem given A1 and A2 are not lipschitz continuous and involve jump discontinuities? because I think I can extend the definition of these functions to discontinuous ones. Thanks $\endgroup$ – Snifkes Sep 28 '18 at 16:38
  • $\begingroup$ The proof above shows that $A_i=-2\sqrt{2}$ almost everywhere (that is, except possibly on a set of measure 0). This is then combined with the continuity restriction to show that this equality must in fact hold everywhere. If you change the continuity condition, you will need to modify this proof - if you run in to trouble, you may consider asking a question about it. $\endgroup$ – KReiser Sep 28 '18 at 17:38

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