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We recall two concepts of differentiability of a mapping. Let $X, Y$ be real Banach spaces and $f: X\rightarrow Y$ be a mapping.

  • $f$ is said to be Frechet differentiable at $\bar{x}$ if there exists a linear continuous operator $\nabla f(\bar{x}): X\rightarrow Y$ such that $ \lim_{x\rightarrow \bar{x}}\frac{f(x)-f(\bar{x})-\nabla f(\bar{x})(x-\bar{x})}{\|x-\bar{x}\|}=0. $

  • $f$ is said to be strictly differentiable at $\bar{x}$ if $f$ is Frechet differentiable at $\bar{x}$ and $ \lim_{\substack{x\rightarrow \bar{x}\\ u\rightarrow \bar{x}}}\frac{f(x)-f(u)-\nabla f(\bar{x})(x-u)}{\|x-u\|}=0. $

It is known that if $f$ is continuously Frechet differentiable in a neighborhooh of $\bar{x}$ then $f$ is strictly differentiable at this point but not vice versa.

I would like to find a mapping $f$ such that $f$ is strictly differentiable at $\bar{x}$ but it is not differentiable at points near $\bar{x}$.

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    If anyone is interested (perhaps having stumbled on this thread at some future time), I posted a lot of references about "strictly differentiable" at ["Strong" derivative of a monotone function](http://math.stackexchange.com/questions/85069/strong-derivative-of-a-monotone-function).2013-01-16

2 Answers 2

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What you call strict differentiability is what Schechter calls strongly differentiable. See chapter 25 of his book.

Section 25.23 proves that

If $f$ is Frechet differentiable on an open set, then $f$ is continuously differentiable if and only if $f$ is strongly differentiable on the same set.

Now to answer your question:

Let $f:\mathbb{R}\to\mathbb{R}$ be given by (below we have $j\in\mathbb{Z}$): $ f(t) = \begin{cases} 2^{2j} + (|t| - 2^{j}) ( 2^{j+2} - 2^j) & |t|\in [2^j, 2^{j+1}) \\ 0 & t = 0 \end{cases} $ Our function $f(t)$ is piecewise linear and continuous, as can be easily checked. It is not differentiable at any $t = 2^j$ since the slope at $(2^j)_+$ and $(2^j)_-$ are different.

One can, however, easily check that $f$ is strongly differentiable at 0 with derivative 0. We see easily that for $x,y$ $ \frac{|f(x) - f(y)|}{|x-y|} \leq \frac{|f(|x|) - f(|y|)}{||x| - |y||} $ so we can assume WLOG $x,y$ has the same sign. Then we have that using the convexity of the construction (the slope increases), that $f(x) - f(y)$ (assuming $x > y$) is at most $3x(x-y)$. And hence as $x,y \to 0$ the limit exists and is 0.

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    Dear Sir. Thank you for your nice and careful answer.2012-10-16
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$f: [-1,1] \rightarrow \mathbb R , \bar x=0$, defined by $f(x):=\begin{cases} x^2 & x=1/k , k\in \mathbb N \\ 0 & x=0\\ \text{linear} & \text{otherwise} \end{cases}$