# Relation between supporting hyperplanes of the graph of a function and supporting hyperplanes of some Cone given by the domain of the function

I am trying to follow the proof of Lemma 9.2 (Certain prerequisite lemma to prove a maximum principle) of the Book of Gilbarg and Trudinger and I am having certain difficulties with an assertion in the proof. It is the following

Let's take a function $u:\Omega\subset\mathbb{R}^n\rightarrow \mathbb{R}$ belonging to the class of $C^2(\Omega)\cap C(\overline{\Omega})$ functions, that takes a positive maximum at a point $y\in \Omega$, and let $k$ be the function whose graph is the cone $K$ with vertex $(y,u(y))$ and base $\partial \Omega \times \{0\}$. Then, the book says that for each supporting hyperplane (Defined later) to $K$, there exists a parallel hyperplane tangent to the graph of $u$. Moreover, as far as I understand, it is not only tangent to the graph of $u$ but also a supporting hyperplane of it. I am utterly lost in even knowing how to start proving this, any hint is welcomed.

Now, the promised definition. We call a supporting hyperplane of the graph of a function $u$ at the point $z$ to an hyperplane given by the graph of a function $f:\Omega\rightarrow \mathbb{R}$ of the form $f(x) := u(z) + p \cdot (x-z)$ such that $u(x)\leq f(x)$ $\forall x\in\Omega$. The $\cdot$ represents the inner product in $\mathbb{R}^n$.

I am purposedly not giving any condition on the set $\Omega$ except posibly being bounded because I don't know if it is necessary for this claim to be truth. I have no problem in assuming it is smooth though because I have to start somewhere.

Apart from that, since I have taken the question from a PDEs book I am tagging it with the PDEs tag. I am not sure which other tags would be appropiate but would thank if any reader comes up with any and tell me in a comment.

Regarding comments: I must add the assumption that $u\leq 0$ on the boundary.

Posibly another necessary assumption: As far as I could prove below (with the help of the answers and comments given here) it may be necessary to ask to the cone $K$ to have a base slightly bigger than $\Omega$. For example $\Omega+B_\varepsilon(0)$. I don't have a counterexample for this assumption to be necessary yet.

• In the case of the horizontal hyperplane, you know $u$ has this as a supporting plane at its maximum, since it is $C^2$. The idea for any other hyperplane is to use the same idea, but with a tilted frame of reference. The fact that the authors are assuming $u \le 0$ on the boundary (which you left out) along with the definition of $K$ should let you show that any such tilted version of $u$ has an interior maximum. – Anthony Carapetis Oct 24 '17 at 5:40
• I followed the ideas of some of the answers and comments below to prove what was needed but liked the idea of proving that the tilted function will have and interior maximum. Unfortunately, I have no idea how to prove this, It seems something that might appear in some books so If you have some reference to read I'd love it. – I.C. Oct 24 '17 at 23:03

Because question didn't specify any point on the graph of $u$ where supporting hyperplane passes through I assume we are free to choose point. (other wise it is not hard always construct a counterexample!)

In later case $K$ does not play any role ! and problem is somehow trivial.

Take any hyper plane $h(x) =a^t x + \alpha$ with normal vector $a \in R^n$ and graph $H \subset R^n \times R$ Consider the shape enclosed with the graph $u$ and $\Omega$, denote it with $S$. Therefore $S \subset R^n \times R$ is a compact set. Now assume $\alpha$ is large enough such that $u \le h$. Thus $dist(H, S)$ is attained on a point on $S$ (a least one point) say $(z , u(z)).$ Now the hyperplan $f(x) := u(z) + a^t \cdot (x-z)$ which is parallel to $h$ and supports $u$ at $(z, u(z))$ and so it is automatically tangent to $u$ too (since $u$ is $C^2$).

In following we show that we choose $f$ such that support $u$ at a point on $\Omega$ not in it's boundary!

Edit To prove claim with assumption $u \leq 0$ on $\partial \Omega ,$ we consider two cases:

Case 1: First Assume $k < u$ on $\Omega$ and $k=u$ on $\partial\Omega.$

Take $h(x) = a^T x + b$ the non-horizontal supporting hyperplane of $K$ at interior (max) point $(z,u(z))=(0,u(0)).$ Hence $k \leq h$ everywhere and $b=u(0) > 0$. It is easy to see that there $w$ close enough to $z=0$ such that $h(w) < u(w).$

Now quite similar previous method by increasing only $b$ to $b + \epsilon$ we can get the parallel hyperplan $f(x) = a^t x + b+ \epsilon$ such that $f$ support $u$ ($u \leq f$) at a point, say $(y,u(y)).$

Claim: $y$ can not be on $\partial \Omega$.

Proof: if not then $f(y) = u(y) = k(y) \leq h(y)$ Therefore

$$a^t y + b + \epsilon \leq a^t + b$$

Which is contradiction, since $\epsilon > 0.$

Case 2: (general case) Let $A$ be the maximal subset of $\Omega$ such that $z=0 \in A$ and $k < u$ on $int A \setminus \{z\}$. " If you are sensitive about existence of $A$ you can apply Zorn Lemma". Moreover, because $u \leq 0$ on $\partial \Omega$ then it is easy to proof that $u = k$ on $\partial A$. Now apply Case 1 on $A$ and get the result.

• If $d(H,S)$ is attained on $\partial \Omega$ (where $S$ has a sharp corner) then this supporting hyperplane will not be tangent to the graph of $u$. This is why you need to use $K$. – Anthony Carapetis Oct 24 '17 at 11:53
• @AnthonyCarapetis So you are saying on presence of $K$ we always can choose the supporting point $(z, u(z))$ on interior of $\Omega$ ?! I doubt it is true... – Red shoes Oct 24 '17 at 18:06
• I posted a community answer trying to fill the details that @Anthony Carapetis comment. I hope I am not breaking any site rules by doing this. If you want to check if it is right then I'll thank it a lot. Edit as necessary. – I.C. Oct 24 '17 at 22:58
• @Red shoes in the last example your cone is inverted and pretty weird, it does not touch zero on the boundary of the domain. If you take $u<0$ in one-dimensional cases the fact that you have an interior point with the slope of your cone is a consequence of mean value theorem as Anthony said. But as I claim in my answer below you use $u$ strictly negative on the boundary to prove that. Otherwise you could end up with a contact point on the boundary. – I.C. Oct 25 '17 at 2:39
• @AnthonyCarapetis it is one of the assumptions that $\max u >0$. – I.C. Oct 25 '17 at 3:12

Fix a hyperplane supporting $K$ at $(p,k(p))$ and let $N=(\nu,\sigma)$ denote its upwards unit normal vector; i.e. the choice with $\sigma > 0.$

Let $h(x,z) = N \cdot (x-p,z-k(p))$ denote the corresponding affine function, which you should think of as the vertical coordinate after tilting to make the hyperplane horizontal.

Claim. The "tilted height" $f(x) = h(x,u(x))$ attains it maximum at some interior point $x_0 \in \Omega$, and thus the graph of $u$ has the desired tangent hyperplane at $(x_0,u(x_0)).$

Proof. For any boundary point $q \in \partial \Omega$ we have \begin{align} f(q) &= (\nu,\sigma) \cdot (q-p,u(q)-k(p)) \\ &= (\nu,\sigma) \cdot (q-p,k(q)-k(p)) + (\nu,\sigma) \cdot (0,u(q)-k(q)) \\ &= h(q,k(q)) + \sigma(u(q)-k(q)). \end{align} Since the hyperplane $h=0$ supports $K$, we know that $h(q,k(q)) \le 0$; so remembering $u|_{\partial \Omega} \le 0$ and $\sigma \ge 0$ we conclude $f(q) \le 0$. Since $K$ is a cone, we know that all its supporting hyperplanes pass through its vertex $(y,u(y))$; so we also have $f(y) = 0$.

As $f$ is continuous up to the boundary and $\bar \Omega$ is compact, we know $f$ attains its maximum somewhere on $\bar \Omega$. We have established that $f$ is non-positive on the boundary but zero at $y$, so the maximum must be attained somewhere in the interior.

• $f$ is not a affine function ! – Red shoes Oct 25 '17 at 8:20
• @Redshoes: I didn't claim it was? If you mean $h$, then it can be written $h(v) = N \cdot v - \nu \cdot p,$ which is clearly affine. – Anthony Carapetis Oct 25 '17 at 9:10
• you firstly called $h$ as supporting hyperplan of $k$ then which function is the parallel affine function to $h$ supporting $u$? in question op called it with $f$. – Red shoes Oct 25 '17 at 16:36
• Oh, my bad for reusing notation. The desired hyperplane supporting $u$ is just the level set of $h$ passing through $(x_0,u(x_0)).$ – Anthony Carapetis Oct 25 '17 at 22:59
• I think you meant $h(x,z) = N \cdot (x-p,z-k(p))$. Moreover $$(\nu,\sigma) \cdot (q,u(q)) = h(q,0) + \sigma u(q)$$ does not hold . Am I right ? – Red shoes Oct 26 '17 at 5:53

I post this as an answer because it was too long for a comment but I am trying to complete what is lacking on @Red shoes answer (now it was deleted, don't know why, but it was a pretty close answer, I'll try to reproduce it later) according to @Anthony Carapetis. I'll post it as a community answer so feel free to edit.

I think that yet there is something wrong with the cone proposed in the book and that it should be slightly modified. I'll introduce one definition of the book and on claim (my supposition here) first and then proceed to the conclusion.

Definition: For an arbitrary function $u\in C^0(\Omega)$ we define the normal mapping, $\chi(y) = \chi_u(y)$ of a point $y\in\Omega$ to be the set \begin{equation*} \chi(y) = \{ p\in \mathbb{R}^n \mid u(x)\leq u(y) + p\cdot (x-y) \quad \forall x\in \Omega\} \end{equation*} and, as usual, $\chi(\Omega)= \cup_{y\in\Omega} \chi(y)$.

With this in mind given the function $k$ whose image is given by the cone with vertex $(z,a)$ and base $B=B_R(z)$. We have, as written in the book, that $\chi_k(B) = B_{a/R}(0)$.

Claim: given two sets $\Omega_1 \subset \Omega_2$ and two functions $k_1, k_2$ whose images are given by the cones of vertice $(z,a)$ with $z\in \Omega_1\subset\Omega_2$ and bases $\Omega_1$ and $\Omega_2$ respectively, we have that $\chi_{k_2}(\Omega_1) \subset \chi_{k_1}(\Omega_1)$. (Imagine two cones one inside the other to get an idea of the relation between the support hyperplanes of each one)

Conclusion: Now I will try to prove that the hyperplanes given in @Red shoes answer are supported in a $z\in \Omega$. For that I take the function $k$ defined in the question post and define an auxiliary function $\tilde{k}$ whose image is given by the cone of vertice $(y,u(y))$ and base $B_R(y)$ with $R=dist(y,\partial \Omega)$. Then, by the claim, we have that $\chi_k(\Omega)\subset \chi_{\tilde{k}}(\Omega)= B_S(0)$ with $S=u(y)/dist(y,\partial \Omega)$. Using this we have that if we take a plane with $p \in B_S(0)$ and let's suppose that it is supported on a point $(z,u(z))$ with $z\in\partial \Omega$ then

$$u(z)\geq u(x) - p\cdot (x-z) \quad \forall x\in \Omega$$ In particular we can take $x=y$ and using the last inequation we have $$|u(z)| \geq |u(y) - p\cdot (y-z)| \geq |u(y)| (1-\frac{|y-z|}{dist(y-z)})\geq 0$$

So here I should conclude that $z$ cannot be in $\partial \Omega$ because $u(z)>0$ but I only have $u(z)\geq 0$ and that is not enough. This is why I said that probably the original cone I should take is one with base $\Omega+\varepsilon"$ in place of just $\Omega$ (That would not be a so bad assumtion to make). Or we should assume $u<0$ on the boundary, but this last posibility is not very usefull because of the context where the question was taken.

• I did , and I added the proof of the statement to my answer. You can check it and let me know if it is not clear for you. @I.C. – Red shoes Oct 26 '17 at 6:09