$f$ is a $k$-Lipschitz continuous function & continuously differentiable $\forall x \in \mathbb{R}^n$. Prove/disprove: $||D_{f}(x)||_{op} \leq k$

Question

EDIT - generalization for vector analysis

for $f: \mathbb{R}^n \to \mathbb{R}^m$, $f$ is continuously differentiable $\forall x \ \in \mathbb{R}^n$, and is $k$-Lipschitz continuous function.

Prove\Disprove: $$\forall x\in \mathbb{R}^n: \ ||D_{f}(x)||_{op} \leq k$$

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previous Version: I was asking the same question but for $f: \mathbb{R} \to \mathbb{R}$

Let $f: \mathbb{R} \to \mathbb{R}$ such that $f$ is continuously differentiable $\forall x \ \in \mathbb{R}$.

Given that $f$ is a $k$-Lipschitz continuous, is it possible to prove that the derivative of $f$ is bounded by $k$?

Prove/disprove: $f'$ is bounded by $k$, as $f$ is a $k$-Lipschitz continuous function ,and continuously differentiable $\forall x \in \mathbb{R}$

The only intuition I have in mind is that:

$$|f(y) - f(x)| = |f'(x)(x-y) + \small{o}(x-y)| \leq k|x-y|$$

Therefore $$ \left|f'(x) + \frac{\small{o}(|x-y)|}{x-y} \right| \leq k$$

But that isn't a sufficent term in order to prove that the derivative of $f$ is bounded by $k$.

Answer 1

We have $$ D_f(x) \xi = \lim_{t\rightarrow 0} \frac{f(x+ t\cdot \xi) - f(x)}{t}.$$ But by the lipschitz condition we get $$ \left \Vert \frac{f(x+ t\cdot \xi) - f(x)}{t} \right\Vert \leq k \Vert \xi \Vert $$ And thus, we have $$ \Vert D_f(x) \xi \Vert \leq k \Vert \xi \Vert $$ implying that $$ \Vert D_f(x) \Vert_{op} \leq k. $$

Answer 2

If $f$ is k -Lipschitz continuous, you have $$ \left| \frac{f(y)-f(x)}{y-x} \right| \le k $$ for all $y \neq x.$ As $f'$ is a limit of such terms, it follows that $|f'| \le k.$

Answer 3

You should isolate instead $f'(x)$, in that case:

$f'(x)=\dfrac{f(x)-f(y)}{x-y}+ \dfrac{o(\vert x-y\vert)}{x-y}= \dfrac{f(x)-f(y)}{x-y}+o(1)$

Then:

$\vert f'(x)\vert \leq \Bigg \vert \dfrac{f(x)-f(y)}{x-y} \Bigg \vert+ \vert o(1)\vert \leq k+ o(1)\overset{y \rightarrow x}{\rightarrow}k$