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Let be $(a_n)\in\ell^2(\mathbb N)$ and consider the mapping $f:\ell^2(\mathbb N)\to\ell^2(\mathbb N)$ given by $$ f\Big((a_n)\Big)=(a_n^n). $$ Question: Is $f$ a Fréchet $C^{\infty}$ function in whole $\ell^2(\mathbb N)$ ?

If the answer for the previous question is no. Is there a non-linear $C^{\infty}$ function, defined in some Banach space that maps a closed bounded set onto a non bounded set ?

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Do you mean "onto" in your last sentence? That is, that the image of the closed bounded set should be an unbounded set? –  Loop Space Aug 17 '10 at 7:04
    
Yes Andrew, I will make the correction. –  Leandro Aug 17 '10 at 7:20
    
In finite-dimensional Banach spaces, closed and bounded is equivalent to compactness (this property actually characterises finite-dimensionality). So in any infinite-dimensional Banach space, closed and bounded is a far weaker property than compactness, so there's no reason why $C^\infty$ functions should have bounded range on a closed, bounded set. I'm sure there are many $C^\infty$ functions with unbounded range on a closed bounded set (it's only the linearity of linear operators which forces bounded range, nothing to do with $C^\infty$). I'll try to think of some on $\ell^2$. –  Zen Harper Aug 17 '10 at 9:42
    
Thanks Zen, I am really curious to know the answer. –  Leandro Aug 17 '10 at 14:27
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4 Answers 4

The behaviour of this function is rather more subtle (and interesting) than suggested by the answers so far. Let me begin with the general remark that the theory of holomorphic functions on Banach spaces (even locally convex spaces) is very well developed---a readable introduction which contains enough material to answer the above questions is a survey article by Nachbin in vol. 79 of the Bulletin of the American Mathematical Society (1973) but a great deal of work has been done since and several monographs have been written on the subject. The function of the question is indeed holomorphic but only on the open unit ball. The problem is that in larger balls there are elements with components larger than $1$ and the powers make them huge. Interestingly, it is holomorphic in weaker senses everywhere, e.g., it is finitely holomorphic, i.e., its restriction to finite dimensional subspaces is holomorphic (in view of Hartogs' theorem it suffices to check this for one-dimensional spaces). It is also holomorphic as a function with values in $\omega$, the product of countable copies of the real line. Very mild smoothness properties combined with finite homomorphicity imply holomorphicity in the strong sense and so these cannot hold for the function in question outside of the unit ball.

With regard to the boundedness question, it is a central fact that holomorphic functions on Banach spaces do not necessarily preserve boundedness---in fact, this behaviour is the exception in a certain sense. For example, a result quoted in the above-mentioned article states that if a closed, bounded subset of a reflexive or separable space is such that each holomorphic mapping is bounded on it, then it is compact.

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Here's an attempt at the second part. I'm much more familiar with the convenient calculus of Frölicher, Kriegl, and Michor than with Fréchet derivatives, but for Hilbert spaces, there shouldn't be too much in it.

Let $\rho \colon \mathbb{R} \to \mathbb{R}$ be a smooth bump function at $0$ with support in $[-1/2,1/2]$. Since $\|e_n - e_m\| = \sqrt{2}$ (for $n \ne m$), if we define $f_n \colon H \to \mathbb{R}$ by $f_n(x) = \rho(|e_n - x|^2)$ then the $f_n$s have disjoint support. Then $f := \sum n f_n \colon H \to \mathbb{R}$ is locally smooth and hence smooth. In addition, $f(e_n) = n$ so $f$ maps the unit ball onto an unbounded set.

(How does that look? I'm well aware that I may have overlooked something really obvious!)


Edit 2012-11-12: I did overlook something that I shouldn't have done. The argument above is not quite correct: that the $f_n$s have disjoint support is not enough to know that the sum $\sum n f_n$ is smooth. For that, I need to know that the $f_n$s have locally finite support. Fortunately, this is true. If $f_n(x) \ne 0$ then $|e_n - x| < 1/4$. Thus for $y \in H$ consider the ball of radius $1/4$ about $y$. If we have $x_1$ and $x_2$ in this ball and $n,m$ such that $f_n(x_1) \ne 0$ and $f_m(x_2) \ne 0$ then $|e_n - e_m| \le |e_n - x_1| + |x_1 - y| + |y - x_2| + |x_2 - e_m| < 1$ whence $n = m$. Hence at most one $f_n$ has support intersecting this ball about $y$ and so the supports of the $f_n$ are locally finite.

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Hi Andrew, I can not see any problem with your argument. But unfortunately, I was not precise in the second question, because I want to understand if there is such example for operators from $\ell^2(\mathbb N)$ to $\ell^2(\mathbb N)$. –  Leandro Aug 17 '10 at 15:19
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I understood that, but you can always compose the function $f$ with one to $\ell^2(\mathbb{N})$: for example, $x \mapsto (f(x),0,0,\dots)$. Or if you want something that looks a little fancier, take the function $x \mapsto \sum n f_n(x) e_n$. Indeed, one can show that there is a function to $\ell^2(\mathbb{N})$ with the properties if and only if there is such a function to $\mathbb{R}$. –  Loop Space Aug 17 '10 at 15:27
    
Of course! Nice example, thank you. –  Leandro Aug 17 '10 at 21:38
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One can compute what the derivative would be $$ (D_a f \cdot h)_n = n (a_n)^{n-1} \cdot h_n. $$ This is just a fancy calculus exercise, to compute what linear functional satisfies $$ \lim\_{t \to 0} \frac{f(a + h t) - f(a) - t \cdot D_a f \cdot h}{t} = 0 $$ So the question boils done to if the multiplication operator $m_a$ with the sequence $n (a_n)^{n-1}$ is bounded on $\ell^2(N)$, and if the map $a \mapsto m_a$ is continuous. I believe that the answer to both questions should be yes.

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It looks easier to check that your function is holomorphic on complex $\ell_2$, from which you can deduce that it is $C^\infty$ (Frechet or Gateaux; it is the same) on real $\ell_2$. Let $f_k(\sum a_n e_n) = a_k^k e_k$, where $(e_k)$ is the unit vector basis. Then the functions $\sum_{k=1}^nf_k$ are holomorphic and converge locally uniformly to your function $f$.

This is either a proof or utter nonsense...

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That last phrase is fantastic! Can I quote you on that? –  Loop Space Aug 17 '10 at 21:08
    
Hi Bill is there a typo in the definition of $f_k$ ? Missing summation ? If not just one more question. You prove that the function is holomorphic in each coordinate, so there is a Hartog version to conclude that the function is holomorphic or it is not necessary too ? –  Leandro Aug 17 '10 at 21:44
    
Sure, Andrew; the phrase is not copyrighted and everyone who knows me would bet heavily on the second possibility. Sorry about the missing summation, Leandro. I added it. IIRC the holomorphic property is preserved by local uniform convergence, but I did not think through a proof or try to look it up. –  Bill Johnson Aug 17 '10 at 22:41
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