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Let $I:=[0,1]$:be an interval,
$\varphi:{I}^{2}\to \mathbb{R}$ : continious on ${I^2}$ and differntiable on $Int({I}^{2})$ ,
$\delta_{(1,c)}:Int(I)\to\mathbb{R}^{2}$ be defined by $t\in Int(I)\mapsto(t,c)\in\mathbb{R}^{2}$

My Question
 For above-mentioned $\varphi$ and $\delta_{(1,c)}$, Is the $\varphi\circ\delta_{(1,0)}(t) = \varphi(t,0)$ a differntiable on $Int(I)$?

If the $\varphi\circ\delta_{(1,0)}$ is differntiable on $Int(I)$, please give me proof or tell me the literature that has description of proof. If not, please give me a counterexample.

Here, $Int({I}^{2})$ is the interior of ${I}^2$ and $Int(I)$ is the interior of ${I}$: these are largest open subset of ${I}^{2}$ and $I$ respectively. For example, $Int(I)=(0,1)=\{t\in\mathbb{R}\ |\ 0<t<1 \}$.
Note that the $(0,1)$ of the previous sentence is an open interval, not a vector. The definition of differentiable is as described in following page:
Are there any functions that are differentiable but not continuously-differentiable?

P.S.
I'm not very good at English, so I'm sorry if I have some impolite or unclear expressions.

Blue Various
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  • What about $\varphi(x,y) = x^{1/2}$? – PhoemueX Sep 01 '19 at 10:25
  • @PhoemueX: Thank you for your comment. I think your $\varphi$ seems not to be a counterexample. Because, $\varphi\circ\delta_{(1,0)}(t) = \varphi(t,0)=\sqrt{t}$ seems differentiable on $(0,1)$ . Even though we change x and y, it seems not to be counterexample. – Blue Various Sep 01 '19 at 10:58
  • @PhoemueX: Perhaps you seem to mention the differentiability at the vector $(0,1)$. However, my question is "About differentiability in the interval (0,1)=Int(I)". – Blue Various Sep 01 '19 at 11:28
  • @Sangchul :Thank you for your comment. Your $varphi$ certainly be a counterexample. (It is not differentiable at $t=1/2$.) Could you post it as an answer? – Blue Various Sep 01 '19 at 11:40
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    Glad it helped, I migrated my comment to an answer. – Sangchul Lee Sep 01 '19 at 11:43

1 Answers1

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(Migrated from comment) Consider the function

$$ \varphi(x, y) = \left( (x - \tfrac{1}{2})^2 + y^2 \right)^{1/2}. $$

Then $\varphi$ is continuous on $I^2$ and differentiable on $\operatorname{Int}(I^2)$, but the map

$$ \varphi \circ \delta_{(1,0)} \quad : \quad t \mapsto \varphi(t, 0) = \left| t - \tfrac{1}{2} \right| $$

is not differentiable on all of $\operatorname{Int}(I)$.

Sangchul Lee
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