$\sup\limits_{y\in [0,1]\setminus\{x\}}\left|\frac{f(x)-f(y)}{x-y}\right|\in \mathbb{R}_+$ continuous?

Asked Mar 20 '20 at 18:24

Active Mar 20 '20 at 18:29

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Is this function $f\in(\mathcal{C}([0,1],\mathbb{R}),\|\cdot\|_\infty)\mapsto B(f)=\sup\limits_{y\in [0,1]\setminus\{x\}}\left|\dfrac{f(x)-f(y)}{x-y}\right|\in \mathbb{R}_+$ continuous?

I think if I reduce this metric space $(\mathcal{C}([0,1],\mathbb{R}),\|\cdot\|_\infty)$ to $E'=\{f\in \mathcal{C}([0,1],\mathbb{R}); \;f(0)=0\}$ then $B$ is a norm.

edited Mar 20 '20 at 18:29

Michael Hardy

asked Mar 20 '20 at 18:24

Stu

1,690

Related: https://math.stackexchange.com/questions/543327/ – Maximilian Janisch Mar 20 '20 at 18:30
Is $x$ just some fixed number? In either case, that supremum need not be finite. – Thorgott Mar 20 '20 at 18:30
the supremum is finite because we have $x\ne y$ – Stu Mar 20 '20 at 18:43
@Stu then the expression is well-defined but still it won't be finite in general. – Sebastian Bechtel Mar 20 '20 at 19:38

$\sup\limits_{y\in [0,1]\setminus\{x\}}\left|\frac{f(x)-f(y)}{x-y}\right|\in \mathbb{R}_+$ continuous?

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