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I am interested in knowing if the following is true:

for any two vectors $v$ and $c$ in a Banach space over the reals : $ \lim_{h \to 0^{+}} \| \dfrac{v}{h} + c \| - \| \dfrac{v}{h}\|$ exists; the motivation is explained below.

This is an exercise from Cartan's Differential Calculus : Let $f$ be a mapping of an interval $[a,b]$ into a Banach space $F$. Let $g(x) = \|f(x)\|$. Show that if $f$ is differentiable on the right at a point $ x \in [a,b]$ then $g$ is also differentiable at this point and $|g^{'}_r(x)| \leq \|f^{'}_r(x)\| $. Here $g_r^{'}$ and $f_r^{'}$ denote the right derivatives of $g$ and $f$ respectively.

So we have to show the existence of $\lim_{h\to{0^{+}}}\dfrac{ \|f(x+h)\| - \|f(x)\|}{h}$ assuming only the existence of $c = \lim_{h \to {0^{+}}} \dfrac{ f(x+h) - f(x)}{h}$ and the continuity of $f$. Assuming this limit exists, the bound is a consequence of $| \dfrac{ \|f(x+h)\| - \|f(x)\|}{h} | \leq \| \dfrac{ f(x+h) - f(x)}{h} \|$ for $h > 0.\ \ $ (1)

I am unable to show to existence of the limit under these general conditions. However it is easy to show the existence of this limit when $f(x) = 0$, $ c = 0 $, or when the norm arises out an inner product.

When $ f(x) = 0 $ then

$\lim_{h\to{0^{+}}}\dfrac{ \|f(x+h)\| - \|f(x)\|}{h} = \lim_{ h \to 0^{+} }\| \dfrac{ f(x+h) - f(x) }{h} \| = \| c \|.$

The above also holds when $c = 0$ from (1).

In case $f(x) \neq 0$ and the norm arises of the inner product, We have for $h > 0$ $ \dfrac{\|f(x+h)\| - \|f(x)\|}{h} =\dfrac{ h\|\dfrac{f(x+h) - f(x)}{h}\|^2 + 2 < \dfrac{f(x+h) - f(x)}{h} , f(x) > }{\|f(x+h)\| + \|f(x)\|}$ which converges to $\dfrac{ < c , f(x) > } { \| f(x) \| } $ as $ h \to 0^+$.

Now for the general case noting that for $ h > 0$ $ \dfrac{ \| f(x+h) \|}{h} - \dfrac{\|f(x)\|}{h} = \| \dfrac{ f(x) }{h} + c + \dfrac{ f(x+h) - f(x) }{h} -c \| - \| \dfrac{f(x)}{h} \| $ and since the previous quantity lies in the interval $ \| \dfrac{ f(x) }{h} + c \| - \| \dfrac{f(x)}{h} \| \pm \|\dfrac{ f(x+h) - f(x) }{h} -c\|$ and $\|\dfrac{ f(x+h) - f(x) }{h} -c\| \to 0$ as $ h \to 0^+$ it is sufficient to show:

for any two vectors $v$ and $c$ in a Banach space : $ \lim_{h \to 0^{+}} \| \dfrac{v}{h} + c \| - \| \dfrac{v}{h}\|$ exists.

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1 Answer 1

up vote 2 down vote accepted

Any convex function of a real variable has one-sided derivatives at every point. Indeed, if $g$ is convex, then $$\frac{g(x+h)-g(x)}{h} \ge \frac{g(x-h)-g(x)}{-h} $$ for all $h>0$. So the LHS is bounded from below, the RHS is bounded from above, and monotonicity with respect to $h$ (which also comes from convexity, by comparing the graph and its secant lines) finishes the story.

Applying the above to the convex function $f(x)=\|v+cx\|$, you will get affirmative answer to your question.

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