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Let $V: L^2(0,1) \to L^2(0,1)$ be the Volterra integration operator: $V(f)(x) := \int_0^x f(t) \, dt$.

Is there a universal function $C(L,\varepsilon) < \infty$ such that the following uniform version of this question holds? For any continuous function $f \in C([0,1])$ satisfying $1 \leq f \leq L$ on the whole segment $[0,1]$, there exist real numbers $a_0, a_1,\ldots, a_M$ such that $$ (1) \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \sum_{k=0}^M \frac{|a_k|}{k!} \leq C(L,\varepsilon) $$ and $$ (2) \quad \quad \quad \quad \quad \quad \Big\| 1 - \sum_{k =0}^M a_k V^k(f) \Big\|_{L^2(0,1)} \leq \varepsilon. $$

(The $a_k/k!$ terms in (1) are natural once one notes that the operator $V^k$ has norm $\leq 1/k!$.)

[A: Yes! As @fedja explained in the comments, a compactness argument applies non-constructively to show that (1) is an automatic consequence of (2). Arguing for contradiction, suppose to the contrary that there is a sequence $f_n$ with $1 \leq f_n \leq L$ that requires an arbitrarily large sum in (1). The set $\{ Vg \mid |g| \leq L \}$ is compact in $C[0,1]$, since it consists of continuously differentiable functions of a bounded $C^1$ norm. Hence, on passing to a subsequence, we may assume that $Vf_n \to f$ uniformly to a continuous function $f \in C[0,1]$, and since all $f_n \geq 1$, this limit function $f$ certainly fulfills a relation (2) with $\varepsilon$ replaced by $\varepsilon / 2$, by the qualitative results cited below the line. If now $n \to \infty$, this is in contradiction with our assumption that all relations (2) for $f_n$ must have a divergent sum in (1). ]

Q.: What could be said about $C(L,0.5)$ as a function of $L$?


The existence of some approximation (2), not necessarily satisfying the uniform condition (1), is known (following M.S. Brodskii) as the unicellularity of the Volterra integration operator, and has (it seems) essentially three fundamentally distinct proofs. One approach (Agmon, Kalisch, Donoghue, Sakhnovic...) is via zero distribution properties of entire functions; indeed the qualitative result (2) turns out to be equivalent to the Titchmarsh convolution theorem. Another proof (Sarason) derives the existence of a relation (2) from Beurling's characterization of the closed invariant subspaces of the unilateral shift of $\ell^2$. A third proof, due to Brodskii and Livsic, is purely functional analytic and based on their spectral resolution theory for abstract non-self adjoint completely continuous operators $V$ of a Hilbert space $H$ having a single point as their spectrum. (The total linear ordering property of the closed invariant subspaces of $V$ holds more generally for all such operators that, in addition, fulfill $\ker(V) \cap \ker(V^*) = \{0\}$ and $\dim\frac{V - V^*}{2i}(H) = 1$.)

To my surprise, I find that apparently none of these three existential schemes of proofs admits an obvious effective variant that says anything about the size of the coefficients $a_k$ used in the representation (2). I would like to know more about the constant $C(L,\varepsilon)$ of this putative statement, particularly as a function of $L$ for a fixed value $\varepsilon$ (is it $O(L)$? $o(L)$?). For a start, I have to understand if there are constructive techniques of functional analysis that would give this refinement, perhaps in combination with one of the existing solutions of (2).

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    $\begingroup$ That would be a bit too much because the norm of $V^k$ in $C[0,0.5]$ is only $\frac 1{k!}2^{-k}$ and you can kill as many low powers as you wish by taking a fast oscillating function (like $\cos Nx$ with huge $N$), so if this is what you really need, you are out of luck. If you can get away with less, let us know what it is. $\endgroup$
    – fedja
    Commented Jul 15, 2018 at 22:47
  • $\begingroup$ @fedja: Thank you for pointing this out: the condition $f(0) = 1$ and $|f| \leq L$ is indeed not the relevant one, in view of these oscillating examples. I was thinking that I should bound $f$ from below somehow, not only at $0$. For my example I can assume $1 \leq f \leq L$ on all of $[0,1]$; I'll edit as this amendment seems to make a difference. In any case, I am curious about any way of quantifying (2), in terms of some such relevant condition about $f$. $\endgroup$ Commented Jul 16, 2018 at 8:31
  • $\begingroup$ Still bad. Take $f(x)=2-\cos Nx$. For the same reason, as before, under condition (1), large $k$ do not matter on $[0,0.5]$ while small $k\ge 1$ give you almost a polynomial, which, since it has bounded degree, cannot catch up with the oscillations of $\cos$ for large enough $N$... $\endgroup$
    – fedja
    Commented Jul 17, 2018 at 16:07
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    $\begingroup$ You are right. For some reason I thought that we were approximating in $C$ rather than in $L^2$. Then some $C(L,\varepsilon)$ exists. Indeed, after the first application of $V$ and passing to the closure, you get a compact family of functions each of which generates the entire space (in $L^2$, but not in $C$). So just find some coefficients for each of them and use the fact that they work in some neighborhood. $\endgroup$
    – fedja
    Commented Jul 17, 2018 at 21:50
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    $\begingroup$ It is used not for compactness but for the statement that every function in the closure of the corresponding family of $Vf$ generates the entire space while in the first counterexample $0$ was in the closure (which is exactly what I used). $\endgroup$
    – fedja
    Commented Jul 18, 2018 at 1:47

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