5
$\begingroup$

Let $f\in C^1(\mathbb T)=C^1(\mathbb R/\mathbb Z)$ be a function such that $$\hat f(k):=\int_{\mathbb T}f(x)e^{-2\pi ikx}\,dx=0,\qquad \forall k\in\{-N+1,\cdots,-1,0,1,\cdots, N-1\}.$$ Do we have $\|f\|_{L^\infty}\leq \frac CN \|f'\|_{L^1}$ for some $C>0$ indpendent of $f$ and $N$?

We have $\|f\|_{L^\infty}\leq \frac1{4N} \|f'\|_{L^\infty}$, see this MSE post for a clever proof. I just wonder, if we change $\|f'\|_{L^\infty}$ to $\|f'\|_{L^1}$, do we have the same inequality?

Maybe the decay $\frac1N$ is too fast to be true. If this is not ture, can we find a $C_N$ such that $$\|f\|_{L^\infty}\leq C_N \|f'\|_{L^1} \qquad \text{and }\ \ \ \ \lim_{N\to\infty}C_N=0?$$

Note that by the fundamental theorem of calculus we obtain $\|f\|_{L^\infty}\leq \|f'\|_{L^1}$. Also, by Cauchy inequality and the Plancherel identity, \begin{align*} \|f\|_{L^\infty}&\leq\sum_{|k|\geq N}|\hat f(k)|=\sum_{|k|\geq N}\frac{\left|\widehat{f'}(k)\right|}{2\pi |k|}\\ &\leq \frac1{2\pi}\left(\sum_{|k|\geq N}\left|\widehat{f'}(k)\right|^2\right)^{1/2}\left(\sum_{|k|\geq N}\frac1{|k|^2}\right)^{1/2}\\ &\leq \frac1{2\pi}\left\|f'\right\|_{L^2}\left(\frac2N\right)^{1/2}=\frac C{\sqrt N}\left\|f'\right\|_{L^2}. \end{align*}

Any help would be appreciated!

$\endgroup$
1
  • $\begingroup$ Note that your considerations above show that you definitely cannot do better than $\frac {C}{\sqrt N}$ since you can choose $f(z)=\sum_{k=N}^M\frac{z^k}{k^2}$ for very large $M$ so $f_{\theta}(e^{i\theta}))=i\sum_{k=N}^M \frac{z^k}{k}$ hence the $L^2$ norm square of $f'$ is then about $1/N$ so its $L^1$ norm being less than the $L^2$ norm is at most of the order $1/\sqrt N$ while $||f||_{\infty}$ is about $1/N$ $\endgroup$
    – Conrad
    Mar 19, 2023 at 14:17

3 Answers 3

10
$\begingroup$

There is no chance. If you could do it, the linear functional $g\mapsto (\int g)(0)$ would have small norm on the corresponding subspace of $L^1$ and, thus, extend to a functional of small norm in the whole $L^1$. Thus, we would have a small in $L^\infty$ function $g$ whose Fourier coefficients are $\frac 1n$ for $|n|>N$, but it would differ only by a trigonometric polynomial from the full series $G(t)=2i\sum_{n\ge 1}\frac {\sin nt}n$, which has a jump discontinuity at $0$. So $g$ cannot eliminate it, and "Whoosh, fly all our hopes away"...

$\endgroup$
9
  • $\begingroup$ not sure exactly what you refer at here in the answer as the OP question was a bit ambiguous; the considerations there and my comment above clearly show that one cannot do better for $L^1$ than what can do for $L^2$ which is easily shown to be $C/\sqrt N$ so definitely not $C/N$; now I suspect that for $L^1$ one can not do $C/\sqrt N$ either, and probably not any other power, but something like $C/\log N$ may be achievable - at least I see no apriori reason $\endgroup$
    – Conrad
    Mar 20, 2023 at 1:34
  • $\begingroup$ @Conrad The constant cannot be made less than the half of the jump of $G$ at $0$ no matter what $N$ is. I thought that was reasonably clear from my explanation. $\endgroup$
    – fedja
    Mar 20, 2023 at 1:41
  • $\begingroup$ Thank you for your answer! However, after some hard thinking and some discussions with my friend, I still cannot follow all the steps... Let $X_0=\{g\in L^1(\mathbb T): \hat g(k)=0, |k|\leq N-1\}$ and define the functional $T_0: X_0\to \mathbb C$ $$T_0g=\int_0^1\left(\int_0^\theta g(\varphi)\,d\varphi\right)\,d\theta,\qquad g\in X_0.$$ Then $|T_0g|\leq C_N\|g\|_{L^1}$. By Hahn-Banach thoerem, we can extend it to a bounded linear functional $T: L^1\to \mathbb C$ with $\|T\|\leq C_N$. Then I'm lost... Why "we would have a small in $L^\infty$ function $g$ whose Fourier coefficients ..."? $\endgroup$
    – Feng
    Mar 20, 2023 at 5:41
  • 2
    $\begingroup$ @Feng Your functional looks weird. I meant the easy one: the value of the obvious antiderivative at $0$, not its integral, i.e., if $g=\sum_n c_n e^{2\pi int}$, then $T(g)=\frac 1{2\pi i}\sum_n\frac{c_n}n$. Then I extend and use the fact that the dual to $L^1$ is $L^\infty$, so $T(g)=\int_0^1 gh$ for some $h\in L^\infty$ with $\|h\|_\infty\le \|T\|$. Now we should have $-\frac 1{2\pi i n}=T(e^{-2\pi nt})=\hat h(n)$ for $|n|>N$ and the rest is as written. $\endgroup$
    – fedja
    Mar 20, 2023 at 13:05
  • $\begingroup$ @fedaj: Thank you so much for this "fedja for dummies" version. It was a neat trick though to use the symbol $g$ for the function $h$ also in the original version. $\endgroup$ Mar 20, 2023 at 13:29
3
$\begingroup$

While @Fedja is an excellent answer, in this particular problem one can construct a polynomial example that shows that indeed one cannot do better than a constant.

Let's take $g(z)=z^N(1+..z^{N-1})^2$, so we have $||g||_{\infty}=N^2$.

But $|g(e^{it})|=(\frac{|\sin (Nt/2)|}{|\sin (t/2)|})^2, -\pi \le t \le \pi$ and $|g(e^{it})| \le c(\frac{|\sin (Nt/2)|}{|t|})^2$ since $|\sin (t/2)| \ge \frac{4}{\pi}|t|, |t| \le \pi$ so one has that $$||g||_1 \le c_1\int_0^{\pi}(\frac{|\sin (Nt/2)|}{|t|})^2dt=c_2N\int_0^{N\pi/2}(\frac{\sin u}{u})^2du \le c_3N$$ since $\int_0^{\infty}(\frac{\sin u}{u})^2du$ converges.

But now one takes $f$ the antiderivative (on the circle) of $g$ so $f(z)=\frac{1}{i}\int_0^z\frac{g(u)du}{u}$ and in particular $||f||_{\infty}=|f(1)|=\int_0^1\frac{g(u)du}{u}$

But for $N \ge 2$ and $1-\frac{1}{N} \le u \le 1$ one has clearly that $u^k \ge u^N \ge 1/4, k \le N$ so $g(u) \ge N^2/64$ so $|f(1)| \ge \int_{1-1/N}^1\frac{g(u)du}{u} \ge c_4N$ hence $||f||_{\infty} \ge c_5||f'||_1$ and manifestly $f$ satisfies the condition that all coefficients up to $N$ are $0$

Edit as per comments - note that $f,g$ are bona-fide polynomials so one can estimate them using their values away from the unit circle, for example on the real axis. $g$ is the typical example of a polynomial of degree of order $N$ ($3N-2$ here) for which its $L^{\infty}$ circle norm is maximal with respect to the $L^1$ norm - it's not hard to show that for polynomials of degree of order $N$, one has $||g||_{\infty} \le cN||g||_1$ ; but we also know that if $g$ divides by $z^N$ and its (circle - so as trigonometric polynomial) antiderivative is $f$ (which is also a regular polynomial of same degree) we have $||g||_{\infty} \ge cN||f||_{\infty}$, so if we find some $g$ that satisfies equality in both the first and second inequalities above we get our counterexample and that is what we proved. Bernstein theorem essentially guarantees equality in the second inequality (but a direct proof here is simple) so we only need $g$ maximal for the infinity norm with respect to the $L^1$ norm as above.

Note that any $z^N(1+...z^{N-1})^k$ for $k \ge 2$ fixed will work in a similar fashion but with worse constants depending on $k$

$\endgroup$
0
2
$\begingroup$

This is an argument inspired by @fedja's answer and gives $C_N=\frac 12$ for every $N$ (which follows also from his proof). We have $$f(0)=\int_{-\pi}^{\pi} f'(\theta) g(\theta)\, d\theta, \quad g(\theta)=-\frac 1\pi \sum_{k=1}^\infty \frac{\sin (k \theta)}{k}.$$ $g$ is piecewise linear: straight line from $(-\pi,0)$ to $(0, 1/2)$ and then from $(0, -1/2)$ to $(\pi,0)$, so that $\|g\|_\infty=1/2$. This gives $|f(0)| \leq \frac 12 \|f'\|_1$ and $C_N \leq 1/2$. To show the converse, for fixed $N$ let $h_\delta$ be the odd reflection of $\frac{1}{2\delta} \chi_{(0, \delta)}$. Then $$h_\delta(\theta)=\frac{1}{\pi \delta}\sum_{k=1}^\infty \frac{1-\cos (k\delta)}{k} \sin (k\theta)=h^1_\delta+h^2_\delta $$ where $h^1_\delta$ is the sum up to $N-1$. Then $|h^1_\delta| \leq CN^2 \delta$ and $h_\delta^2$ has non vanishing Fourier coefficients only for $|k| \geq N$. Moreover, $\|h_\delta^2\|_1 =1+O(\delta)$ and $\int_{-\pi}^\pi h_\delta^2 g= -\frac 12 +O(\delta)$. If $f_\delta'=-h_\delta^2$, then $f_\delta(0) \geq \left (\frac 12 +O(\delta)\right )\|f'_\delta\|_1$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.