# How to prove this partial derivative?

Consider $u:\mathbf{R}\times\omega\rightarrow\mathbf{R}$, where $\omega\subset\mathbf{R}^{n-1}$ is a bounded domain. For each $y\in\omega$ and each $\lambda>0$, consider $y^\lambda=(x,2\lambda-y_1,y_2,...,y_{n-1})$ the reflection of $y$ in the plane $\{y_1=\lambda\}$. Suppose that $$w_\lambda(x,y)=u(x,y)-u(x,y^\lambda)<0.$$ If $y_1>0$ then $$\frac{\partial u}{\partial y_1}\leq0.$$

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Are you sure this is the right formulation of the question? I think there is no function u with this property because $w_{y_1}(x,y)=u(x,y)-u(x,y^{y_1})=0$ – Dominik Dec 4 '12 at 17:37
Do you assume that $\lambda$ is such that $u(x,y^\lambda)$ is defined? We must have $y^\lambda \in \omega$. – Zach L. Dec 4 '12 at 19:38
Zach, $\lambda\in(0,\sup_\omega y_1)$ then $u(x,y^\lambda)$ is well defined. Dominik, when you take $\lambda=y_1$, you have $w_{y_1}=0$, it is ok. I know that this consequence is true, but i don't know how to argue. I accept ideas. Thanks. – José Carlos Dec 5 '12 at 5:57

Claim: If $w_\lambda(x,y)\leq0$, then $\frac{\partial u}{\partial y_1}(x,\lambda,y_2,\dots,y_n)\leq0$.

Suppose the theorem is not true, that is, $w_\lambda(x,y)\leq0$ and $\frac{\partial u}{\partial y_1}(\bar x,\bar y)=\ell>0$ for some $(\bar x,\bar y)\in \mathbb R^n$. Then, by the definition of the partial derivative, $$\ell=\lim_{\delta\to0} \frac{u(\bar x,\bar y+\delta)-u(\bar x,\bar y-\delta)}{2\delta},$$ (where $\bar y+\delta=(\bar y_1+\delta,\bar y_2,\dots,\bar y_{n-1})$ above). Thus we have for some $\delta$ such that $\epsilon<\ell$, $$0<2\delta(\ell-\epsilon)<u(\bar x,\bar y+\delta)-u(\bar x,\bar y-\delta)=w_{\bar y_1}(\bar x,\bar y+\delta).$$

This is a contradiction, so for any $\lambda$ such that $w_\lambda(x,y)\leq0$, we have $\frac{\partial u}{\partial y_1}(x,\lambda,y_2,\dots,y_n)\leq0$. Note that this conclusion is not true for points $y_1\neq\lambda$: If $$u(x,y_1)=\frac{\lambda-y_1}{(y_1-\lambda)^2+1}\Rightarrow\frac{\partial u}{\partial y_1}=\frac{(y_1-\lambda)^2-1}{(1+(y_1-\lambda)^2)^2},$$ with $\omega=\mathbb R$ and $\lambda>0$, then $w_\lambda(x,y_1)=(\lambda-y_1)\frac2{(y_1-\lambda)^2+1}<0$ if $y_1>\lambda$, but $\frac{\partial u}{\partial y_1}>0$ for $y_1>\lambda+1>0$.

Please tell me if I'm missing anything, but it seems as if the problem is a lot simpler than it sounds at face value (the multidimensional stuff doesn't matter at all in this proof). I'm also assuming that your constraint $w_\lambda(x,y)\leq0$ only applies for $y_1\geq\lambda\Rightarrow y_1\geq y_1^\lambda$, because $w_\lambda(x,y)=-w_\lambda(x,y^\lambda)$, so the constraint would make no sense otherwise unless $w_\lambda(x,y)=0$ everywhere.

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Hi! Im sorry for the delay. I think that I solved the problem. This hipothesys implies in the symmetry of the function in $y_1$, in other words, $$u(x,y_1,...,y_{n-1})=u(x,-y_1,...,y_{n-1}).$$ Then, $$u(x,y_1-2\lambda,y_2,...,y_{n-1})-u(x,y)=u(x,2\lambda-y_1,y_2,...,y_{n-1})-u(x‌​,y)>0.$$ Since $\lambda>0$, $$\frac{u(x,y_1-2\lambda,y_2,...,y_{n-1})-u(x,y)}{-2\lambda}<0.$$ Taking $t=-2\lambda$, $$\frac{\partial u}{\partial y_1}(x,y)=\frac{\partial u}{\partial y_1^-}(x,y)=\lim_{t\To0^-}\frac{u(x,y_1+t,y_2,...,y_{n-1})-u(x,y)}{t}\leq0.$$ Finally $$\frac{\partial u}{\partial y_1}\leq0.$$ – José Carlos Dec 11 '12 at 2:56

Is the first inequality(apparently there must be $\leq$ instead <) true for each $\lambda\in\ (0,sup_\omega\ y_1)$ or just for certain value of $\lambda$ ? If this is true for each value of $\lambda$ then it seems to me that the second inequality turns to equality

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One may forget about the variables $x$, $y_2$, $\ldots$, $y_{n-1}$. So we are given an open interval $\Omega\subset{\mathbb R}$ and a differentiable function $$u:\quad \Omega\to{\mathbb R},\qquad y\mapsto u(y)\ .$$ For each $\lambda\in \Omega$ the expression $$w_\lambda(y):=u(y)-u(2\lambda -y)$$ is defined in a neighborhood of $y=\lambda$, and it is assumed that $w_\lambda(y)\leq0$ wherever the left side makes sense.

I claim that under these circumstances the function $u$ is constant in $\Omega$.

Proof. Consider a fixed $y_0\in\Omega$, and write $y=y_0+h$ for small $|h|$. Choose $\lambda:=y_0$. Then $$w_\lambda(y)=u(y_0+h)-u(y_0-h)=\bigl(2u'(y_0)+o(1)\bigr) h\qquad (h\to 0)\ .$$ As the left side is $\leq0$ for all $h$ it follows that necessarily $u'(y_0)=0$, and since this is true for all $y_0\in\Omega$, the function $u$ has to be constant.

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