Example f bounded, f' unbounded and Lipschitz I need help with a proof or counterexample. I have the conjecture that there is no differentiable function $f:\mathbb{R}^d\mapsto\mathbb{R}$ (or even $f:\mathbb{R}\mapsto\mathbb{R}$) for which:
\begin{align*}
\text{1. $f$ is bounded } \forall x\in\mathbb{R}^d:\;&\vert f(x)\vert\leq c_0\\
\text{2. the gradient $\nabla f$ is Lipschitz } \forall x,y\in\mathbb{R}^d:\;&\Vert \nabla f(x)-\nabla f(y)\Vert \leq c_1\Vert x-y\Vert\\
\text{3. the gradient $\nabla f$ is unbounded } \forall c<\infty\; \exists x\in\mathbb{R}^d:\;&\Vert \nabla f(x)\Vert \geq c
\end{align*}
Such a function must be only once differentiable.
For twice differentiable functions the Lipschitz derivative 2. implies a bounded second derivative:
Lipschitz-constant gradient implies bounded eigenvalues on Hessian.
A bounded second derivative together with f being bounded 1. implies that the derivative is bounded $\not$3.:
Is there a bounded function $f$ with $f'$ unbounded and $f''$ bounded?
Assumption 2. and 3. can be satisfied, when 1. is violated, e.g. $f:\mathbb{R}\rightarrow\mathbb{R}:x\mapsto \frac{1}{2}x^2$ has an unbounded, Lipschitz derivative $f'(x)=x$ (and bounded second derivative $f''(x)=1$).
Assumption 1. and 3. can be satisfied, when 2. is violated, e.g. $f:\mathbb{R}\rightarrow\mathbb{R}:x\mapsto sin(xsin(x))$ is a bounded, differentiable functions with unbounded derivative that is not Lipschitz. Can the graph of a bounded function ever have an unbounded derivative?.
 A: Let $f:\mathbb{R}^d\rightarrow\mathbb{R}$ be a differentiable function for which:
\begin{align*}
\text{1. $f$ is bounded by $c_1$} \forall x\in\mathbb{R}^d:\;&\vert f(x)\vert \leq c_1\\
\text{2. the gradient is $c_2$-Lipschitz } \forall x,y\in\mathbb{R}^d:\;&\Vert \nabla f(x)-\nabla f(y)\Vert \leq c_2\Vert x-y\Vert
\end{align*}
Then the gradient is bounded by $\forall x\in\mathbb{R}^d:\;\Vert \nabla f(x)\Vert\leq 2\sqrt{c_1c_2}$.
The differentiability of $f$ and the Fundamental Theorem for Line Integrals provide for any $x,h\in\mathbb{R}^d$:
\begin{align*}
\langle\nabla f(x),h\rangle+R(h)=f(y)-f(x)
=\int_0^1\langle\nabla f(x+\lambda h),h\rangle\; d\lambda
\end{align*}
The remainder $R(h):\mathbb{R}^d\rightarrow\mathbb{R}$ can be bounded for any $h\in\mathbb{R}^d$ using the Cauchy Schwarz inequatility and Lipschitz assumption:
\begin{align*}
\vert R(h)\vert &= \Big\vert \int_0^1\langle\nabla f(x+\lambda h),h\rangle\; d\lambda - \langle\nabla f(x),h\rangle\Big\vert\\
&= \Big\vert \int_0^1\langle\nabla f(x+\lambda h)-\nabla f(x),h\rangle \;d\lambda\\
&\leq  \int_0^1  \Vert \nabla f(x+\lambda h)-\nabla f(x) \Vert \Vert h \Vert \;d\lambda \\
&\leq  \int_0^1  \lambda c_2\Vert h \Vert^2 \;d\lambda 
\leq \frac{c_2}{2}\Vert h\Vert^2
\end{align*}
Rearranging and using that $f$ is bounded for all $x,h\in\mathbb{R}^d$:
\begin{equation*}
\vert \langle\nabla f(x),h\rangle \vert = \vert f(x+h)-f(x) - R(h) \vert \leq 2c_1+ \frac{1}{2}c_2\Vert h\Vert^2
\end{equation*}
Choosing $h=2\sqrt{\frac{c_1}{c_2}}\frac{\nabla f(x)}{\Vert \nabla f(x)\Vert}$ concludes for all $x\in\mathbb{R}^d$:
\begin{align*}
2\sqrt{\frac{c_1}{c_2}} \Vert \nabla f(x)\Vert &\leq  2c_1 +2c_1=4c_1\\
\Vert \nabla f(x)\Vert&\leq 2\sqrt{c_1c_2}
\end{align*}
