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Let $K\subset \mathbb{R}^n$ be a closed convex cone with a nonempty interior. Let $f:K\to \mathbb{R}$ be a continuously differentiable function satisfying $\|\nabla f\|_\infty<\infty$ (if needed, you can further assume $\|\nabla f\|_{\text{Lip}}<\infty$).

Note that the interior of $K$ being nonempty implies that the linear span of all vectors in $K$ is $\mathbb{R}^n$. In this sense, we say that $K$ spans $\mathbb{R}^n$.

We also assume that $\nabla f$ exists on $ K$ including the boundary in the following sense: for any fixed $x \in K$, $f(y)-f(x) = (y-x)\cdot \nabla f(x) + o(|y-x|)$ for all $y\in K$ as $y\to x$. (Since $K$ spans $\mathbb{R}^n$, $\nabla f(x)$ is unqiue.)

But, we do not assume $f$ is bounded and $f$ is allowed to grow linearly.

Question: Is there a continuously differentiable function $F:\mathbb{R}^n\to\mathbb{R}$ such that $F|_K =f$ and $\|\nabla F\|_\infty<\infty$?

I looked at a few versions of Whitney's theorem, but either there is no control on $\nabla F$ or $\|f\|_\infty<\infty$ is required.

It would be great if there were published results to cite directly. If not, my naive attempt to prove this is to make those covering cubes in the proof of Whitney's theorems to scale along the cone.

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  • $\begingroup$ $K$ is closed and convex and $K$ spans $\mathbb{R}^n$ implies that $K=\mathbb{R}^n$. Also if $f$ is continuously differentiable on the closed $K$, then certainly $\nabla f$ is defined on $\partial K$ since $\partial K\subset K$. Then in what sense $\nabla f$ is defined on $\partial K$? Your question requires numerous clarifications. $\endgroup$
    – Piotr Hajlasz
    Commented Dec 11, 2022 at 14:00
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    $\begingroup$ I clarified the meanings. Thanks for pointing them out! @PiotrHajlasz $\endgroup$
    – mathoverflow_user
    Commented Dec 11, 2022 at 14:17

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$\newcommand\R{\mathbb R}$According to this answer, the distance function $d\colon K^c\to\R$ to $K$ given by the formula $$d(x):=\min\{|y-x|\colon y\in K\}$$ for $x\in K^c:=\R^n\setminus K$ is differentiable, where $|\cdot|$ is the Euclidean norm.

So, I think the formula $$F(x):=\int_{C_x}[f(y)+\nabla f(y)\cdot(x-y)]\,dy\Big/\int_{C_x}\,dy$$ for $x\in K^c$, where $C_x:=\{y\in K\colon |y-x|\le2d(x)\}$, will provide a desired extension of $f$. However, I have not checked the details.

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  • $\begingroup$ Thank you! But, the link to ``the answer'' is not working. Can you fix it? $\endgroup$
    – mathoverflow_user
    Commented Dec 11, 2022 at 15:49
  • $\begingroup$ @mathoverflow_user : Thank you for your comment. The link is now fixed, I think. $\endgroup$
    – Iosif Pinelis
    Commented Dec 11, 2022 at 16:01
  • $\begingroup$ I can't verifying $\nabla F(x) = \nabla f(x)$ for $x\in \partial K$, which should hold because $K$ spans $\mathbb{R}^n$. Since $F(z) - F(x) = ⨍_{C_z} (y-x)\cdot \nabla f(x)+o(|y-x|) d y$, comparing this with $(z-x)\cdot \nabla f(x)$ we want to make sure that $⨍_{C_z}(y-z)\cdot\nabla f(x) d y=o(|z-x|)$. This is where I don't think it works in general. $\endgroup$
    – mathoverflow_user
    Commented Dec 11, 2022 at 17:52
  • $\begingroup$ @mathoverflow_user : Do you want to look at the modified suggestion? $\endgroup$
    – Iosif Pinelis
    Commented Dec 11, 2022 at 18:55
  • $\begingroup$ Yes, I can verify that $F$ is differentiable on $K$. But I can't verify that on $K^c$. Maybe it's easy since it's away from the boundary, but I am not experienced enough to see it right away. Could you please elaborate on the differentiability on $K^c$? Thank you! $\endgroup$
    – mathoverflow_user
    Commented Dec 12, 2022 at 11:27

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