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Let us consider the integral operator $T:L^2(0,1)\to [0,\infty)$ such that for all $k\in L^2(0,1)$, $$ T(k)=\int_0^1 k_t^2e^{-2 \int_0^t k_s d s} d t. $$

Is the operator $T$ coercive in the $L^2$ sense, i.e., $ T(k)\to \infty$ as $|k|_{L^2}\to \infty$?

Let us assume without loss of generality that $k_t\ge 0$ for all $t$. If we employ the rough estimate: $$ T(k)\ge \int_0^1 k_t^2 d t e^{-\int_0^1 2k_s\, d s}\ge |k|^2_{L^2} e^{-2|k|_{L^2}}, $$ then it seems to indicate that the functional is not coercive, as $|k|_{L^2} \exp(- 2|k|_{L^2})$ tends to zero as $|k|_{L^2}\to \infty$.

However, the above inequality is a quite poor lower bound. In fact, if we consider the sequence of functions $(k^n)_{n\in \mathbb{N}}$ where for each $n\in \mathbb{N}$, $k^n=\alpha_n/2$ on $[0,1/n]$ and zero otherwise, then $|k^n|^2_{L^2}=\alpha_n^2//4n$, and a direct computation shows that $$ T(k^n)\sim \alpha_n (1-e^{-\alpha_n/n}), \quad n\in \mathbb{N}, $$ and hence $\lim_{n\to \infty}\alpha_n^2/n=\infty$ indeed implies $\lim_{n\to \infty} T(k^n) = \infty$.

I feel I am closer to prove that the claim is false. Consider $k_t=(1-t)^{-1}$ for all $t\in (0,1)$. Note that $k\not \in L^1(0,1)$. A direct computation shows that $ \int_0^t k_s d s =-\ln (1-t)$ and hence $$ T(k)=\int_0^1 (1-t)^{-2}e^{2\ln (1-t)} d t=1. $$ So if I could find a sequence of $(k^n)_{n\in \mathbb{N}}$ approaching $k$ with $\sup_{n\in \mathbb{N}} T(k^n)<\infty$ then I may get a counterexample.

John
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  • By $k_s$ do you mean $k(s)$? In that case is $k_s^2$ supposed to be $k_t^2$ as the integration variable is $t$? – user23571113 Sep 27 '22 at 20:46
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    @user23571113 Thank you for pointing out my typo. I have corrected the error in my question. – John Sep 27 '22 at 21:03

1 Answers1

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The claim is false. Here is a counterexample. For each $\epsilon>0$, consider $k^\epsilon_t=(1+\epsilon-t)^{-1}$ for all $t\in (0,1)$. Note that $k^\epsilon \in L^2(0,1)$, with $\int_0^1 (k^\epsilon_t)^2 d t=\frac{1}{\epsilon(1+\epsilon)}$. In particular $|k^\epsilon|_{L^2}\to \infty$ as $\epsilon\to 0$.

A direct computation shows that $ \int_0^t k^\epsilon_s d s =\ln(1+\epsilon)-\ln (1+\epsilon-t)$ and hence $$ T(k^\epsilon)=\int_0^1 (1+\epsilon-t)^{-2}e^{2(\ln (1+\epsilon-t)-\ln(1+\epsilon))} d t=\frac{1}{(1+\epsilon)^2}. $$ So $T(k^\epsilon)$ remains finite as $\epsilon \to 0$. A similar conterexample has been given in Proposition 2.4 of this paper.

John
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