An unpleasant measure theory/functional analysis problem

Question

I am currently taking a functional analysis course, and at the moment every student on the course is stumped by a specific question. We're looking at the bounded linear map $$ \varphi_n:L^1\left([0,1]\right)\to \mathbb{R},\quad \varphi_n(f) = \int^1_0 f(x)(n\sin(n^2 x)) dx,\quad\|\varphi_n\|=n. $$

We are attempting to prove the existence of an $f\in L^1$ with respect to the Lebesgue measure, so that $\lim_{n\rightarrow \infty} |\varphi_n(f)|\rightarrow \infty$, and from an earlier assignment we know that this function cannot be continuously differentiable. However, nobody has been able to make even the slightest progress. I, and all of us would appreciate a hint or a sketch of proof!

Answer 1

I think that for $\alpha\in (\tfrac12,1)$ the function $$ f(x) = \begin{cases} x^{-\alpha} &\text{if}\;x\in(0,1) \\ 0 &\text{if}\;x\not\in (0,1) \end{cases} $$ satisfies $$ \varphi_n(f)\to\infty \qquad(n\to\infty). $$ The proof requires some estimations because it is probably rather difficult to calculate the integral exactly.

Let $n\geq 3$ be fixed. For $k\geq 0$, we define $$ I_k := \int_{k\pi n^{-2}}^{(k+1)\pi n^{-2}} f(x)(n\sin(n^2 x)) \mathrm dx. $$ It is clear that $\varphi_n(f) = \sum_{k\geq 0} I_k$ and that $I_k=0$ for large $k$. It is also easy to see that $I_k\geq0$ if $k$ is even and $I_k\leq0$ if $k$ is odd.

Due to the monotonicity of $f$ it can be shown that $ I_k+I_{k+1}\geq 0 $ if $k$ is even.

Let us calculate a lower estimate for $I_0+I_1$. Again, for $k=0,1,2,3$ we define the integrals $$ J_k := \int_{k\pi n^{-2}/2}^{(k+1)\pi n^{-2}/2} f(x)(n |\sin(n^2 x)|) \mathrm dx. $$ It is clear that $I_0=J_0+J_1$ and $I_1=-J_2-J_3$. For $k=0,1,2,3$ we know that $f(x)\geq ((k+1)\pi n^{-2}/2)^{-\alpha}$. Therefore we have the lower estimate $$ J_k \geq \int_0^{\pi n^{-2}/2}((k+1)\pi n^{-2}/2)^{-\alpha} n \sin(n^2 x)\mathrm dx = \dots = ((k+1)\pi/2)^{-\alpha} n^{2\alpha-1}. $$ On the other hand, for $k=1,2,3$ we know that $f(x)\leq (k\pi n^{-2}/2)^{-\alpha}$ Therefore we have the upper estimate $$ J_k \leq \int_0^{\pi n^{-2}/2}(k\pi n^{-2}/2)^{-\alpha} n \sin(n^2 x)\mathrm dx = \dots = (k\pi/2)^{-\alpha} n^{2\alpha-1}. $$ Using these estimates it can be seen that $$ I_0-I_1 = J_0+J_1-J_2-J_3 \geq n^{2\alpha-1} ( (\pi/2)^{-\alpha} +(2\pi/2)^{-\alpha}-(2\pi/2)^{-\alpha}-(3\pi/2)^{-\alpha}) = C n^{2\alpha-1} $$ for some constant $C>0$.

To summarize, we have $$ \varphi_n(f) \geq I_0+I_1 \geq C n^{2\alpha-1} \to \infty \quad (n\to\infty). $$

Answer 2

Uniform boundedness principle will show that a function $f$ must exist such that $|\phi_n(f)|$ is unbounded. Else $\phi_n(f)$ is bounded for each $f$ in a Banach space, and so $\|\phi_n\|$ would be a bounded sequence of functionals, which it is not. I do not know however how an explicit example would look.

Answer 3

@supinf showed us a nice set of examples. I think the proof can be simplified. So suppose $1/2<p<1$ and let $f(x)=x^{-p}.$ Let $x=y/n^2$ to see

$$\tag 1 n\int^1_0 x^{-p}\sin(n^2 x)\, dx = n^{2p-1}\int_0^{n^2} y^{-p}\sin(y)\, dy.$$

If we show $\int_0^{\infty} y^{-p}\sin(y)\, dy >0,$ then it will follow that $(1)\to \infty$ on the order of $n^{2p-1}.$ But this integral equals

$$\sum_{k=0}^{\infty}\int_0^{\pi}\sin y\left(\frac{1}{(2\pi k+y)^p} - \frac{1}{(2\pi k+\pi +y)^p}\right)\,dy.$$

Each of the integrands here is positive on $(0,\pi),$ hence so is the sum, and we're done.