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Let $H$ be a Hilbert space. I would like to establish the following property (if it is true): for every $F \in C^0([0,1];H)$, for every $x_0 \in [0,1]$ (or a.e. $x \in (0,1)$), there exists a sequence $\{\varphi_n\}_{n \geq 1} \subset L^2((0,1);\mathbb{R})$ such that $$ \left\{ \begin{array}{c} \int_0^1 F(x)\varphi_n(x) \, dx \longrightarrow F(x_0) \mbox{ as } n \to +\infty, \\ \exists C>0, \quad {\|\varphi_n\|}_{L^2(0,1)} \leq C, \quad \forall n \geq 1. \end{array} \right. $$

Note that this is the $L^2$-norm that we want to be bounded, not the $L^1$-norm. This is my main problem.

Remark: I consider vector-valued functions $F$ so that one can not divide and take the constant sequence $\varphi_n(x)=F(x_0)/\int F$ if $\int F \neq 0$.

1) My first try was to take $$\varphi_n(x)=n 1_{\left[x_0,x_0+\frac{1}{n}\right]}(x),$$ (assuming $x_0<1$). Then, the first property holds since $F$ is continuous near $x_0$ (set $\epsilon=\frac{1}{n}$ for a better view), but $\|\varphi_n\|_{L^2(0,1)}=\sqrt{n}$ is not bounded.

2) My second try is to view this as a convolution. Let $\bar{F}$ be any continuous extension of $F$ to $\mathbb{R}$. Set $\varphi_n(x)=\rho_n(x_0-x)$ where $\rho_n \in C^{\infty}_0(\mathbb{R})$ is an approximate of the identity. The classical choice is $$\rho_n(x)=\frac{n}{{\|\rho\|}_{L^1(\mathbb{R})}} \rho(nx),$$ where $\rho \in C^{\infty}_0(\mathbb{R})$ is such that $\mathrm{supp} \, \rho \subset B(0,1)$, $\rho \geq 0$ and $\rho \neq 0$.

Assume that $0<x_0<1$ and let $n$ large enough so that $0<-\frac{1}{n}+x_0$ and $\frac{1}{n}+x_0<1$. Then, since $\mathrm{supp} \, \rho_n \subset (-\frac{1}{n},\frac{1}{n})$, so that $\mathrm{supp} \, \rho_n(x_0-\cdot) \subset (-\frac{1}{n}+x_0,\frac{1}{n}+x_0) \subset (0,1)$, we have $$(\rho_n \ast \bar{F}) (x_0) =\int_{\mathbb{R}} \rho_n(x_0-x) \bar{F}(x) \, dx =\int_0^1 \varphi_n(x)F(x) \, dx .$$ Since $\bar{F} \in C^0(\mathbb{R})$ we know that $ \rho_n \ast \bar{F} \rightarrow \bar{F}$ uniformly on every compact. This gives the first property. However, a computation shows that, again, the norm is not bounded: $$\|\varphi_n\|_{L^2(0,1)} ={\|{\rho_n|}}_{L^2\left(-\frac{1}{n},\frac{1}{n}\right)}=\left( \frac{{\|\rho\|}_{L^2(\mathbb{R})}}{{\|\rho\|}_{L^1(\mathbb{R})}}\right) \sqrt{n}.$$

Any idea ?

Edit:

I am interested in vector-valued functions since I know that it is true for $H=\mathbb{R}$. Indeed, assume that the properties hold. Then, since ${\{\varphi_n\}}_{n \geq 1}$ is bounded in the Hilbert space $L^2(0,1)$, there exists subsequence, still denoted ${\{\varphi_n\}}_{n \geq 1}$, converging weakly to some $\varphi \in L^2(0,1)$. Passing to the limit $n \to +\infty$ in the first property we obtain that $\varphi$ necessarily satisfies $$\int_0^1 F(x)\varphi(x) \, dx=F(x_0). \qquad (1).$$ Conversely, if (1) holds that we can take $\varphi_n=\varphi$. Thus, it is equivalent to prove the existence of $\varphi \in L^2(0,1)$ such that (1) holds. If $F \equiv 0$, this is trivially solved taking any $\varphi$. Assume then that $F \not\equiv 0$. Then, there exists $\psi \in L^2(0,1)$ such that $I=\int_0^1 F(x)\psi(x) \, dx \neq 0$. Multiplying this equation by $F(x_0)/I$ we obtain (1) for $\varphi(x)=\psi(x) F(x_0)/I$.

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A more typical question would be to ask for $(\phi_n)$ such that $\int F\phi_n\to F(x_0)$ for every $F$. That's impossible with $||\phi_n||_2$ bounded, even if we consider just scalar-valued $F$: It follows easily that $\hat\phi_n(\xi)\to e^{-ix_0\xi}$ pointwise, so Fatou's lemma shows that $||\phi_n||_2=||\hat\phi_n||_2\to\infty$.

In the vector-valued context I suspect that it's impossible even for fixed $F$. For example, let $$F(x)=\chi_{(x,1)}\in L^2;$$I suspect that if $\int_0^1 F(x)\phi_n(x)\to F(0)=\chi_{(0,1)}$ in $L^2$ then $||\phi_n||_2\to\infty$. (Because since $F$ has orthogonal increments it seems the only way this can happen is if $\lim_{n\to\infty}\int_0^\delta\phi_n=1$ for every $\delta>0$, which implies $||\phi_n||_2\to\infty$.)

Yes: For $0<\lambda<1$ let $$h_\lambda=\lambda^{-1/2}\chi_{(0,\lambda)}.$$Since $||h_\lambda||_2=1$, if $||\phi_n||_2\le C$ (and assuming wlog that $\phi_n$ iis real-valued) we have $\newcommand{\ip}[2]{\langle #1,#2\rangle}$ $$\begin{aligned}||\chi_{(0,1)}-\int_0^1 F(x)\phi_n(x)\,dx||_2 &\ge\ip{h_\lambda}{\chi_{(0,1)}-\int_0^1F(x)\phi_n(x)\,dx} \\&=\lambda^{1/2}-\int_0^1\ip{h_\lambda}{F(x)}\phi_n(x)\,dx \\&=\lambda^{1/2}-\lambda^{-1/2}\int_0^\lambda(\lambda-x)\phi_n(x)\,dx \\&\ge \lambda^{1/2}-C\lambda^{-1/2}\left(\int_0^\lambda(\lambda-x)^2\,dx\right)^{1/2} \\&= \lambda^{1/2}-C\lambda/\sqrt3.\end{aligned}$$

There exists $\lambda\in(0,1)$ such that $\lambda^{1/2}-C\lambda/\sqrt3>0$; hence $||\chi_{(0,1)}-\int_0^1 F(x)\phi_n(x)\,dx||_2$ does not tend to zero.

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  • $\begingroup$ Thank you for your answer. I don't understand your first paragraph since we are working on a bounded domain. For $F:x \mapsto e^{-ix\xi} \in C^0([0,1])$, the first property gives $\hat{\phi}_n(\xi)=\int_{\mathbb{R}} \overline{\phi_n(x)} e^{-ix\xi} \, dx \rightarrow e^{-ix_0 \xi}$ for every $\xi \in [0,1]$ (where $\overline{\phi_n}$ is the extension of $\phi_n$ by zero outside $(0,1)$). Since $|\hat{\phi}_n(\xi)| \leq \|\phi_n\|_{L^2(0,1)} \leq C$ for every $\xi \in (0,1)$, we can apply the dominated convergence theorem. $\endgroup$ – perturbation Feb 8 '18 at 17:13
  • $\begingroup$ Thank you for this nice counter-example in the second paragraph. However I still wonder if the property is valid for a.e. $x_0 \in [0,1]$ ? You provided a counter-example with $x_0=0$ but if $x_0>0$ then, for the same $h_\lambda$, you can not let $\lambda \to 0$ to obtain a contradiction. Indeed, your leading term now vanishes for $\lambda$ small enough: $${\langle h_\lambda, F(x_0)\rangle}_{L^2(0,1)}=\lambda^{-1/2} \int_0^{\lambda} 1_{[x_0,1]}(s) \, ds=0, \quad \forall \lambda<x_0.$$ $\endgroup$ – perturbation Feb 8 '18 at 17:13
  • $\begingroup$ @perturbation Regarding the first paragraph: ?????? The function $x\mapsto e^{-ix\xi}$ is continuous on $[0,1]$ for every $\xi\in\Bbb R$; this has nothing to do with whether or not $\xi\in[0,1]$. So $|\hat\phi_n(\xi)|\to1$ for every $\xi\in\Bbb R$, hence $||\phi_n||_2=||\hat\phi_n||_2\to\infty$. $\endgroup$ – David C. Ullrich Feb 8 '18 at 17:43
  • $\begingroup$ @perturbation Regarding almost-everywhere convergence for that example: I have no idea. Of course the argument I gave applies only to the case $x_0=0$. Try $h_\lambda=\chi_{(x_0,x_0+\lambda)}$ and see what happens... $\endgroup$ – David C. Ullrich Feb 8 '18 at 17:46
  • $\begingroup$ Yes you are right ! Sorry, I mixed the variables in my mind... $\endgroup$ – perturbation Feb 8 '18 at 17:51

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