Suppose $\phi:\mathbb{R}^n\to \mathbb{R}$ is a $C^1$ function and fix $x \in \mathbb{R}^n$.
For every $t \in (0,+\infty)$ let $y_t:[0,+\infty)\to \mathbb{R}^n$ be a continuous function such that $y_t(0)=x$ and $$\left\lVert y_t (s)-x\right\rVert\leq Ms\qquad s\in[0,+\infty)$$ for a fixed $M>0$ (which does not depend on $t$) .
1) Define $g:(0,+\infty)\to \mathbb{R}^+$ such that
$$g(t)=\sup_{s \in [0,t]}\left\lVert \nabla\phi(y_t(s))-\nabla\phi(x)\right\rVert$$
for every $t\in (0,+\infty)$, where $\nabla \phi$ denotes the gradient of $\phi$.
Is it true that $\lim_{t\to 0^+}g(t)=0$?
2)Define $h:(0,+\infty)\to \mathbb{R}^+$ such that
$$h(t)=\sup_{s \in [0,t]}\left\lVert \nabla\phi(y_t(s))\right\rVert.$$
Is it true that $\lim_{t\to 0^+}h(t)=\left\lVert \nabla\phi(x)\right\rVert$?
PERSONAL ATTEMPT
1) Fix $\epsilon>0$. Since $\nabla \phi$ is continuous at $x$ there exist $\overline{\delta}>0$ such that for every $z \in \mathbb{R}^n$ with $\left\lVert z-x\right\rVert\ < \overline{\delta}$ one has $\left\lVert \nabla\phi(z)-\nabla\phi(x)\right\rVert < \frac{\epsilon}{2}$. Let $\delta:=\frac{\overline{\delta}}{M}$.
Now fix $0<t<\delta$. For every $s \in [0,t]$ one has $0\leq s\leq t<\delta$ and so $\left\lVert y_t(s)-x\right\rVert \leq Ms<M\delta=\overline{\delta}$, so $\left\lVert \nabla\phi(y_t(s))-\nabla\phi(x)\right\rVert<\frac{\epsilon}{2}< \epsilon$. Then $$g(t)=\sup_{s \in [0,t]}\left\lVert \nabla\phi(y_t(s))-\nabla\phi(x)\right\rVert \leq \frac{\epsilon}{2}<\epsilon.$$
2) Fix $\epsilon>0$. Since 1) holds, then there exist $\delta>0$ s.t. for every $0<t<\delta$ one has $0\leq g(t)<\epsilon$. Let $0<t<\delta$ fixed. Then $$\sup_{s \in [0,t]}\left\lVert \nabla\phi(y_t(s))\right\rVert-\left\lVert\nabla\phi(x)\right\rVert=\sup_{s \in [0,t]}\{ \left\lVert \nabla\phi(y_t(s))\right\rVert-\left\lVert\nabla\phi(x)\right\rVert\}\leq\sup_{s \in [0,t]}\{\left\lVert \nabla\phi(y_t(s))-\nabla\phi(x)\right\rVert\},$$ since $$\left\lVert \nabla\phi(y_t(s))\right\rVert-\left\lVert\nabla\phi(x)\right\rVert\leq\left\lVert \nabla\phi(y_t(s))-\nabla\phi(x)\right\rVert\qquad s \in [0,t].$$ Conversely $$\left\lVert \nabla\phi(x)\right\rVert-\sup_{s \in [0,t]}\left\lVert \nabla\phi(y_t(s))\right\rVert\leq 0\leq \sup_{s \in [0,t]}\{\left\lVert \nabla\phi(y_t(s))-\nabla\phi(x)\right\rVert\}, $$ hence $$|h(t)-\left\lVert\nabla\phi(x)\right\rVert|=|\sup_{s \in [0,t]}\left\lVert \nabla\phi(y_t(s))\right\rVert-\left\lVert\nabla\phi(x)\right\rVert|\leq \sup_{s \in [0,t]}\{\left\lVert \nabla\phi(y_t(s))-\nabla\phi(x)\right\rVert\}=g(t)<\epsilon.$$
Hope my reasonings are correct. Any feedback would be really appreciated.