User angelo lucia - MathOverflow

Convolution in $\ell_p$ when $0
2013-04-03T16:31:53Z

Background

It is known that given real sequences $a = (a_n)_{n \in \mathbb Z} \in \ell_p$ and $b = (b_n)_{n \in \mathbb Z} \in \ell_q$, their convolution defined as $$ a * b (n) = \sum_{k \in \mathbb Z} a_{n-k} b_k $$ is in $\ell_r$ if $1 \le p, q < \infty$ and $\frac 1 r = \frac 1 p + \frac 1 q -1 $.

Question

What happens when $0 < p, q < 1$? Obviously, since $a$ and $b$ are in $\ell_1$, their convolution $a * b$ is in $\ell_1$. Can we say better, i.e. $a*b$ is in $\ell_r$ for some $r < 1$?

Answer by Angelo Lucia for Usefulness of Frechet versus Gateaux differentiability or something in between.

2011-08-12T14:06:38Z

I know this question has been inactive for such a long time, but I will try to add something.

Dieudonne in Treatise on analysis defines an "intermediate" concept of differentiability between Fréchet and Gâteaux ones called quasi-differentiability. Using your notation, $V$ is quasi-differentiable at $x$ if $V \circ g$ is differentiable at 0, for all $g : [0,1] \to L$ which are continuous, right-differentiable in $0$ , and $g(0) = x$ .

(I am not sure about it, but i think you could caracterize quasi-differentiability in terms of uniform convergence on compact sets, i.e. the limit $ \lim_{t\to 0} \frac{V(x+tu) - V(x)}{t}$ is uniform on compact sets.)

If $L$ is finite dimensional, then quasi-differentiability implies Fréchet differentiability. If $V$ is Lipschitz, then Gâteaux differentiability implies quasi-differentiability.

Nice thing about quasi-differentiability is that the theorem of differentiation of composite functions holds (wich is not the case with Gâteaux differentiability). Moreover, if $F$ is quasi-differentiable and $G$ is Gâteaux differentiable, then $F \circ G$ is G-differentiable.

Unfortunately, it's possible to construct a function $F : L^2 \to L^2$ which is everywhere quasi-differentiable, its differential is a right inverse (thus injective), but is nowhere locally injective. This is not possible if $F$ is Fréchet differentiable.

Answer by Angelo Lucia for A good book of functional analysis

2011-08-09T07:54:57Z

Apart from the classics already mentioned (Yosida, Brezis, Rudin), a good book of functional analysis that I think is suitable not only as a reference but also for self-study, is Fabian, Habala et al. Functional Analysis and Infinite-Dimensional Geometry. It has a lot of nice exercises, it's less abstract than the usual book and provides a lot of "concrete" theorems.

And I'm not sure about it, but I heard there is a spanish translation (the original is of course in english).

Equivalent metrics on Fréchet spaces and Lipschitz maps

2011-08-06T10:47:08Z

Lipschitz maps are defined over metric space as maps $f:(X,d_X) \to (Y,d_Y)$ such that $$ d\left( f(x),f(x^\prime) \right)_Y \le k d(x,x^\prime)_X \ \forall x,x^\prime \in X, $$ where $k$ is a positive constant. We usually say that $f$ is a contraction if $k<1$ .

It is well know that a different equivalent metric on $X$ does not preserve contractions, i.e. a map can be a contraction with respect to a metric but not with respect to an equivalent one.

In the Banach space setting, where the spaces $X$ and $Y$ are endowed with a norm defining the topology, there is a somehow "canonical" distance given by $$ d(x,x^\prime) = \lVert x-x^\prime \rVert .$$ With this distance, Lipschitz maps can be characterized as maps satisfying, for some $k>0$ $$ {\left\lVert f(x) - f(x^\prime) \right\rVert}_Y \le k {\left \lVert x-x^\prime \right \rVert}_X \ \forall x, x^\prime \in X. $$ It is obvious that, if $f$ satisfies the above relation, then is a $k$ -lipschitz map.

In the Fréchet space setting, the topology is defined by a countable family of semi-norms $({\lVert\cdot\rVert}_n)$ . The classical example of metric inducing the same topology is given by $$ d(x,x^\prime) = \sum_{n=0}^\infty {2^{-n}} \frac{{\lVert x-x^\prime\rVert}_n}{1+{\lVert x-x^\prime\rVert}_n} . $$

In analogy with the Banach case, I would like to characterize (at least some) Lipschtiz maps between Fréchet spaces as maps satisfying $$ {\left\lVert f(x) - f(x^\prime) \right\rVert}_n \le k {\left \lVert x-x^\prime \right \rVert}_n \ \forall x, x^\prime \in X,\ \forall n \in \mathbb{N}. $$ Again, maps satisfying the last equation are Lipschitz maps with respect to the metric defined above, but the Lipschitz constant is not $k$ anymore, and in particular contraction with respect to the semi-norms (i.e. maps satisfying the last equation with $k<1$ ) are not contraction with respect to the metric.

Are there equivalent distances on $X$ and $Y$ such that every contraction with respect to the semi-norms is a contraction with respect with the new distance? If this is not possibile for every contraction, is it possible for a specific one?

Comment by Angelo Lucia

2013-04-09T13:36:43Z

Just one one: should it be "Fubini", not "Funini"?

Comment by Angelo Lucia

2013-04-09T10:42:51Z

Looks like the optimal exponent is exactly max$(p,q)$. To see this, consider two strictly positive sequences: $a_n, b_n >0$. Then trivially $$ a*b(n) = \sum_{k\in \mathbb Z} a_{n-k} b_k \ge a_n b_0 ;$$ and similarly $a*b(n) \ge b_n a_0$. Then $a*b(n)$ cannot go to zero faster than the slower between $a$ and $b$. @Davide: if you want, you should post an answer.

Comment by Angelo Lucia

2011-08-08T08:39:48Z

...but it still has a unique fixed point, right?

Comment by Angelo Lucia

2011-08-07T21:29:41Z

I meant $\sum 2^{-n} (\left\lVert x-y \right\rVert \wedge k)$ on $Y$ .

Comment by Angelo Lucia

2011-08-07T20:03:10Z

I understand the idea, but I would say that if I take $ \sum 2^{-n} ({\left\lVert x-y\right\rVert} \wedge 1)$ on $X$ , I need $\sum 2^{-n} ({\left\lVert x-y\right\rVert} \wedge 1/k)$ on $Y$ . Anyway... what about maps from $X$ into himself? Are maps $f:X\toX$ satisfying $\left\lVert f(x) -f(y)\right\rVert \le k \left \lVert x-y \right \rVert$ , with $k<1$ , distance-contraction (and thus have a unique fixed point)? maybe this is a different question...