Hilbert Inequality

Inequalities of the form

\displaystyle \sum_{m=1}^{\infty} \sum_{n=1}^{\infty} K(m,n)a_{m} b_{n}<C_K \left||a_{n}\right\|_{2}\left\lVert b_{n}\right\|_2

and more generally, for p>1, \frac{1}{p}+ \frac{1}{q} =1 ,

\displaystyle \sum_{m=1}^{\infty} \sum_{n=1}^{\infty} K(m,n)a_{m} b_{n}<C_K \left||a_{n}\right\|_{p}\left\lVert b_{n}\right\|_q

are called Hilbert’s inequalities (Hardy-Hilbert Inequalities).

This can be viewed in terms of matrices/operators. Let T: l^2 \to l^2 be defined as

\displaystyle x=(x_n) \to Ax,\quad (T x)_m =\sum_{n=1}^{\infty}K(m,n)x_n

We are looking for bounds of the form \displaystyle \langle y, Tx \rangle \le C \left||x\right\|_{2}\left\lVert y\right\|_2 . By Cauchy-Schwarz, this is equivalent to having

\| Tx\|_2 \le C\|x\|_2

Thus we are trying show the operator T: l^2 \to l^2 is bounded and the best constant C is the operator norm.

We start with the case where the kernel K(m,n) =\frac{1}{m+n}.

Theorem 1 :

\displaystyle \sum_{m=1}^{\infty} \sum_{n=1}^{\infty} \frac{a_{m} b_{n}}{m+n}<C\left(\sum_{m=1}^{\infty} |a_{m}|^{2}\right)^{\frac{1}{2}}\left(\sum_{n=1}^{\infty} |b_{n}|^{2}\right)^{\frac{1}{2}}

The above inequality holds with C=\pi and no smaller constant. It was first proved by Hermann Weyl but with C=2\pi, the best constant was later obtained by Schur.

More generally we have the integral version:

\displaystyle \int_{0}^{\infty} \int_{0}^{\infty} \frac{f(x) g(y)}{x+y} d x d y<\pi \left||f\right\|_{2}\left\lVert g\right\|_2

This l^2 result can be generalized to l^p inequality

\displaystyle \sum_{m=1}^{\infty} \sum_{n=1}^{\infty} \frac{a_{m} b_{n}}{m+n}< \pi \csc \left(\frac{\pi}{p}\right)\left||a_{n}\right\|_{p}\left\lVert b_{n}\right\|_q

Proofs:

Toeplitz’s method:

Write \frac{1}{n} in terms of exponentials (Fourier series) using

\displaystyle \frac{1}{2 \pi} \int_{0}^{2 \pi}(t-\pi) e^{i n t} d t=\frac{1}{i n}

Now

\displaystyle \sum_{m=1}^{\infty} \sum_{n=1}^{\infty} \frac{a_{m} b_{n}}{m+n} =\frac{i}{2 \pi} \int_{0}^{2 \pi}(t-\pi) \sum_{m=1}^{\infty} a_{m} e^{i m t} \sum_{n=1}^{\infty} b_{n} e^{i n t} d t

Now applying Cauchy-Schwarz we get

\displaystyle \le \frac{1}{2\pi} \left(\int_{0}^{2\pi} (t-\pi)^2 \left|\sum_{m=1}^{\infty} a_{m} e^{i m t}\right|^2\right)^{1/2} \left(\int_{0}^{2\pi} \left|\sum_{n=1}^{\infty} b_{n} e^{i n t}\right|^2\right)^{1/2}

\displaystyle \le \frac{1}{2\pi} \left(\int_{0}^{2\pi} \pi^2 \left|\sum_{m=1}^{\infty} a_{m} e^{i m t}\right|^2\right)^{1/2} \left(\int_{0}^{2\pi} \left|\sum_{n=1}^{\infty} b_{n} e^{i n t}\right|^2\right)^{1/2}

\displaystyle =\pi\left(\sum_{m=1}^{\infty} a_{m}^{2}\right)^{\frac{1}{2}}\left(\sum_{n=1}^{\infty} b_{n}^{2}\right)^{\frac{1}{2}}

where the step is Parseval- just open the square and integrate- only diagonal terms m_1=m_2, n_1=n_2 remain, non-diagonal terms integrate to zero.

\square

Tightness: How do we know this is optimal?

Consider a_{n}(t)=b_{n}(t)=n^{-\frac{1}{2}-t} and see what happens as t \to 0.

We can see that the LHS of the inequality is

\displaystyle \sum_{m=1}^{\infty} \sum_{n=1}^{\infty} \frac{1}{n^{\frac{1}{2}+t}} \frac{1}{m^{\frac{1}{2}+t}} \frac{1}{m+n} \sim \int_{1}^{\infty} \int_{1}^{\infty} \frac{1}{x^{\frac{1}{2}+\epsilon}} \frac{1}{y^{\frac{1}{2}+\epsilon}} \frac{1}{x+y} d x d y \sim \frac{\pi}{2 t}

and RHS is \sim \frac{C}{2t},~ therefore C has to be at least \pi.

We crucially used the l^2 features when we executed Cauchy-Schwarz and Parseval. How do we prove the l^p version?

Method of Compensating Difficulties:

To start of we think of the LHS as sum over (m,n) of the quantities \frac{a_{m}}{\sqrt{m+n}}\quad and \frac{b_{n}}{\sqrt{m+n}}

Applying Cauchy-Schwarz we get

\displaystyle \sum_{m=1}^{\infty} \sum_{n=1}^{\infty} \frac{a_{m} b_{n}}{m+n} \le \sum_{m=1}^{\infty} \sum_{n=1}^{\infty} \frac{a_{m}^{2}}{m+n} \sum_{n=1}^{\infty} \sum_{m=1}^{\infty} \frac{b_{n}^{2}}{m+n}

But these sums are not convergent. So we multiply the factor \left(\frac{m}{n}\right)^{\lambda} to the first sequence and a compensate it by multiplying the second factor with \left(\frac{n}{m}\right)^{\lambda}

That is we apply Cauch-Schwarz to \displaystyle \frac{a_{m}}{\sqrt{m+n}}\left(\frac{m}{n}\right)^{\alpha} \quad and \displaystyle \frac{b_{n}}{\sqrt{m+n}}\left(\frac{n}{m}\right)^{\alpha}

to get

\displaystyle \sum_{m=1}^{\infty} \sum_{n=1}^{\infty} \frac{a_{m} b_{n}}{m+n}\le \sum_{m=1}^{\infty} \sum_{n=1}^{\infty} \frac{a_{m}^{2}}{m+n}\left(\frac{m}{n}\right)^{2 \alpha} \sum_{n=1}^{\infty} \sum_{m=1}^{\infty} \frac{b_{n}^{2}}{m+n}\left(\frac{n}{m}\right)^{2 \alpha}

Now

\displaystyle \sum_{m=1}^{\infty} \sum_{n=1}^{\infty} \frac{a_{m}^{2}}{m+n}\left(\frac{m}{n}\right)^{2 \alpha}=\sum_{m=1}^{\infty} a_{m}^{2} \sum_{n=1}^{\infty} \frac{1}{m+n}\left(\frac{m}{n}\right)^{2 \alpha}

We have

\displaystyle \sum_{n=1}^{\infty} \frac{1}{m+n}\left(\frac{m}{n}\right)^{2 \alpha} \leq \int_{0}^{\infty} \frac{1}{m+x} \frac{m^{2 \alpha}}{x^{2 \alpha}} d x=\int_{0}^{\infty} \frac{1}{(1+t)} \frac{1}{t^{2 \alpha}} d t

Therefore we get

\displaystyle \sum_{m=1}^{\infty} \sum_{n=1}^{\infty} \frac{a_{m} b_{n}}{m+n}< \int_{0}^{\infty} \frac{1}{(1+t)} \frac{1}{t^{2 \alpha}} d t \left(\sum_{m=1}^{\infty} a_{m}^{2}\right)^{\frac{1}{2}}\left(\sum_{n=1}^{\infty} b_{n}^{2}\right)^{\frac{1}{2}}

The integral \displaystyle \int_{0}^{\infty} \frac{1}{(1+t)} \frac{1}{t^{2 \alpha}} d t converges for 0<\alpha< 1/2 and evaluates to

\displaystyle \int_{0}^{\infty} \frac{1}{(1+t)} \frac{1}{t^{2 \alpha}} d t =\frac{\pi}{\sin 2 \pi \alpha}

And at \alpha =1/4, we have the best choice which is \pi.

This method allows us to get do better- for instance if the sum is restricted to n \le N. Note that the

\displaystyle \sum_{m=1}^{\infty} a_{m}^{2} \sum_{n=1}^{\infty} \frac{1}{m+n}\left(\frac{m}{n}\right)^{2 \alpha} by \displaystyle\sum_{m=1}^{\infty} a_{m}^{2} \int_{0}^{\infty} \frac{1}{(1+t)} \frac{1}{t^{2 \alpha}} d t

Take \alpha =1/4– this is where we got the best constant.

\displaystyle \sum_{n=1}^{\infty} \frac{1}{m+n}\left(\frac{m}{n}\right)^{1/2} \le \pi -\frac{c}{\sqrt n}

So we have the improved bound

\displaystyle \left(\sum_{m=1}^{\infty}\left(\pi-\frac{c}{m^{1 / 2}}\right) a_{m}^{2}\right)^{\frac{1}{2}} \left(\sum_{n=1}^{\infty}\left(\pi-\frac{c}{n^{1 / 2}}\right) b_{n}^{2}\right)^{\frac{1}{2}}.

So if the sequences are restricted to n \le N, m\le M, we get the improved bound

\displaystyle \sqrt{(\pi -c/N^2)(\pi-c/M^2)} \left(\sum_{m=1}^{M}a_{m}^{2}\right)^{\frac{1}{2}} \left(\sum_{n=1}^{N} b_{n}^{2}\right)^{\frac{1}{2}}.

We can carryout all of the above with l^p: Instead of Cauchy-Schwarz we apply Holder with

\displaystyle \frac{a_{m}}{(m+n)^{1/p}}\left(\frac{m}{n}\right)^{\alpha} \quad and \displaystyle \frac{b_{n}}{(m+n)^{1/q}}\left(\frac{n}{m}\right)^{\alpha}

We get

\displaystyle \sum_{m=1}^{\infty} \sum_{n=1}^{\infty} \frac{a_{m} b_{n}}{m+n}\le \left(\sum_{m=1}^{\infty} |a_m|^p \sum_{n=1}^{\infty}\frac{1}{m+n} \left(\frac{m^{\alpha p}}{n^{\alpha p}}\right) \right)^{1/p} \left(\sum_{n=1}^{\infty} |b_n|^q \sum_{n=1}^{\infty}\frac{1}{m+n} \left(\frac{n^{\alpha q}}{m^{\alpha q}}\right) \right)^{1/q}

We bound

\displaystyle \sum_{n=1}^{\infty}\frac{1}{m+n} \left(\frac{m^{\alpha p}}{n^{\beta p}}\right) \le \int_{0}^{\infty} \frac{1}{(1+t)} \frac{1}{t^{p \alpha}} d t =\frac{\pi}{\sin \pi \alpha p}

So we have the bound

\left(\frac{\pi}{\sin (\pi \alpha p)}\right)^{1/p}\left(\frac{\pi}{\sin (\pi \alpha q)}\right)^{1/q} \|a_n\|_p \|b_n\|_q

The best constant \frac{\pi}{\sin (\frac{\pi}{p})}=\frac{\pi}{\sin (\frac{\pi}{q})} is obtained when \alpha =\frac{1}{pq}.

\displaystyle K(m,n) =\frac{1}{(m+n)^{\tau}}

In this case we get

\displaystyle \sum_{n=1}^{\infty} \sum_{m=1}^{\infty} \frac{a_{n} b_{m}}{(n+m)^{\tau}}<\left\{\pi \csc \left(\frac{\pi(q-1)}{\tau q}\right)\right\}^{\tau}\left\|a_{n}\right\|_{p}\left\|b_{n}\right\|_{q}

Max version:


\displaystyle \sum_{m=1}^{\infty} \sum_{n=1}^{\infty} \frac{a_{m} b_{n}}{\max (m, n)}<4 \left\|a_{n}\right\|_{2}\left\| b_{n}\right\|_2

4 is the best constant here.

Homogenous Kernels:

K(\lambda x, \lambda y) =\lambda^{s}K(x, y)

Homogenous of degree s=-1 :

The integral version is

\displaystyle \int_{0}^{\infty} \int_{0}^{\infty} K(x, y) f(x) g(y) d x d y=\int_{0}^{\infty} f(x) d x \int_{0}^{\infty} x K(x, x u) g(x u) d u
\displaystyle =\int_{0}^{\infty} f(x) d x \int_{0}^{\infty} K(1, u) g(x u) d u
\displaystyle =\int_{0}^{\infty} K(1, u) d u \int_{0}^{\infty} f(x) g(x u) d x
\displaystyle \le \int_{0}^{\infty} |K(1, u)| d u\left(\int_{0}^{\infty} f(x)^{2} d x\right)^{1 / 2}\left(\int_{0}^{\infty} g(x u)^{2} d x\right)^{1 / 2}
\displaystyle =\left(\int_{0}^{\infty} |K(1, u)| u^{-1 / 2} d u\right) \|f\|_2 \|g\|_2

\displaystyle C=\left(\int_{0}^{\infty} |K(1, u)| u^{-1 / 2} d u\right) =\left(\int_{0}^{\infty} |K(u, 1)| u^{-1 / 2} d u\right)

is the best possible constant a non-negative kernel K(x,y) homogenous of degree -1.

For the discrete case, we cannot apply some of the rescaling steps in the above argument but if

\displaystyle K(1, y) y^{-1 / 2} and \displaystyle K(x, 1) x^{-1 / 2} are both non-negative decreasing functions, we have

\displaystyle \sum_{m, n=1}^{\infty} K(m, n) a_{m} b_{n}=\sum_{m, n=1}^{\infty}\left(a_{m} K(m, n)^{1 / 2}\left(\frac{m}{n}\right)^{1 / 4}\right)\left(b_{n} K(m, n)^{1 / 2}\left(\frac{n}{m}\right)^{1 / 4}\right)

by Cauchy-Schwarz we get

\displaystyle \le \left(\sum_{m=1}^{\infty} a_{m}^{2} \sum_{n=1}^{\infty} K(m, n)\left(\frac{m}{n}\right)^{1 / 2}\right)^{1 / 2}\left(\sum_{n=1}^{\infty} b_{n}^{2} \sum_{m=1}^{\infty} K(m, n)\left(\frac{n}{m}\right)^{1 / 2}\right)^{1 / 2}

Now

\displaystyle \sum_{m=1}^{\infty} a_{m}^{2} \sum_{n=1}^{\infty} K(m, n)\left(\frac{m}{n}\right)^{1 / 2}=\sum_{m=1}^{\infty} a_{m}^{2} \sum_{n=1}^{\infty} K\left(1, \frac{n}{m}\right)\left(\frac{n}{m}\right)^{-1 / 2} 1 / m

and by the monotonocity we have

\displaystyle \sum_{r=1}^{\infty} K(1, r / m)(r / m)^{-1 / 2} 1 / m<\int_{0}^{\infty} K(1, y) y^{-1 / 2} d y

Therefore we get

\displaystyle \sum_{m,n=1}^{\infty} K(m,n)a_{m} b_{n}<C_K \left||a_{n}\right\|_{2}\left\lVert b_{n}\right\|_2

with the constant

\displaystyle C= =\left(\int_{0}^{\infty} K(1, u) u^{-1 / 2} d u\right) =\left(\int_{0}^{\infty} K(u, 1) u^{-1 / 2} d u\right)

Note that the above proof contains/matches the proof by compensating difficulties in the case of \frac{1}{m+n} because

K(x,y) =\frac{1}{x+y} is non-negative and homogenous of degree -1.

The following can be obtained by the same ideas.

\displaystyle \sum_{n=1}^{\infty} \sum_{m=1}^{\infty} \frac{\ln \left(\frac{m}{n}\right) a_{m} b_{n}}{m-n}<\pi^{2} \left||a_{n}\right\|_{2}\left\lVert b_{n}\right\|_2

because

\displaystyle \int_{0}^{\infty} \frac{-\ln u}{1-u} u^{-1 / 2} d u =\pi^2

\displaystyle\sum_{n=1}^{\infty} \sum_{m=1}^{\infty} \frac{|\ln (m / n)| a_{m} b_{n}}{\max (m, n)}<8\left||a_{n}\right\|_{2}\left\lVert b_{n}\right\|_2

because

\displaystyle \int_{0}^{\infty} \frac{|\ln u|}{\max(1, u)} u^{-1 / 2} d u =8

The max version


\displaystyle \sum_{m=1}^{\infty} \sum_{n=1}^{\infty} \frac{a_{m} b_{n}}{\max (m, n)}<4 \left\|a_{n}\right\|_{2}\left\| b_{n}\right\|_2

because

\displaystyle \int_{0}^{\infty} \frac{1}{\max(1, u)} u^{-1 / 2} d u =4

Under the same assumptions, we can get l^p versions:

\displaystyle \int_{0}^{\infty} \int_{0}^{\infty} K(x, y) f(x) g(y) d x d y <C_{K,p,q} \|f\|_p \|g\|_q

\displaystyle \sum_{m=1}^{\infty} \sum_{n=1}^{\infty} K(m,n)a_{m} b_{n}<C_{K,p,q} \left||a_{n}\right\|_{p}\left\lVert b_{n}\right\|_q

where the constant is \displaystyle C_{K, p, q}=\int_{0}^{\infty} K(x, 1) x^{-1 / p} d x =\int_{0}^{\infty} K(1, y) y^{-1 / q} d y.

Harder Hilbert Inequality: \displaystyle K(m,n) =\frac{1}{m-n}, ~~ \text{if} ~~m \neq n, \quad 0~~ \text{otherwise}

\displaystyle \mathop{\sum\sum}_{m,n\ge 1, m \neq n}^{\infty} \frac{a_{m} b_{n}}{m-n}<\pi\left(\sum_{m=1}^{\infty} a_{m}^{2}\right)^{\frac{1}{2}}\left(\sum_{n=1}^{\infty} b_{n}^{2}\right)^{\frac{1}{2}}

Using the Fourier formula for \frac{1}{m-n} (Toeplitz method), we see that LHS is bounded by

\displaystyle \frac{1}{2\pi} \int_{0}^{2\pi} |t-\pi| \left| \sum_{m=1}^{\infty} a_m e^{i m t}\right| \left|\sum_{n=1}^{\infty} b_n e^{-i n t} \right|

which by Cauch-Schwarz is bounded by

\displaystyle \le \pi \|a_n\|_2\|b_n\|_2

Note that we are bounding |t-\pi| by the constant \pi on the interval [0, 2\pi]. If we also include the variation of this function (and the fact that most of the l^2 mass comes from t \approx 0 \mod 2\pi), you save a little from the variation of |t-\pi| near t \approx 0 2\pi and get

\displaystyle \mathop{\sum\sum}_{m,n\ge 1, m \neq n}^{N} \frac{a_{m} b_{n}}{m-n} \le \left(\pi-\frac{c}{N}\right) \|a_n\|_2\|b_n\|_2

We derive the above estimate with c=\pi using some auxillary functions/operators.

For \displaystyle K_1(m,n) =\csc (\frac{\pi(m-n)}{R}), we see that operator norm is exactly R-1

In fact the eigenvalues are (1-N)i, \pm (3-N) i, \cdots (2k-1-N)i,\cdots \quad k=1, 2,\cdots N. Therefore the operator norm , the size of the largest eigenvalue is R-1.

To verify that just observe that e^{\frac{(2 k-1) n\pi i}{N}} are the eigenvectors with eigenvalues

\lambda_k = -\sum_{n=1}^{N} e^{\frac{(2 k-1) n \pi i}{N}} \csc \frac{\pi n}{N} = (2k-1-N)i

Thus we have

\displaystyle \left|\sum_{m \neq n}^{N} a_n b_m \csc \frac{\pi(m-n)}{N}\right| \leq(N-1) \|a_n\|_2 \| b_n\|_2

(In fact given m,n \le N, applying this inequality for N_1 > N large with a_n =0, N <n\le N_1, and taking the limit N_1 \to \infty we recover

\displaystyle \mathop{\sum\sum}_{m,n\ge 1, m \neq n}^{N} \frac{a_{m} b_{n}}{m-n}<\pi\left(\sum_{m=1}^{\infty} a_{m}^{2}\right)^{\frac{1}{2}}\left(\sum_{n=1}^{\infty} b_{n}^{2}\right)^{\frac{1}{2}}

because \frac{\sin x}{x} \to 1 \text{ as } x \to 0

But we want to get a better constant than $late x \pi$ for this finite version.

Now we write

\displaystyle \frac{1}{m-n} =\csc(\frac{\pi (m-n)}{N}) \frac{\sin (\frac{\pi (m-n)}{N})}{m-n}

\displaystyle =\csc(\frac{\pi (m-n)}{N}) \frac{\pi}{N} \int_{-1 / 2}^{1 / 2} e\left(\frac{(m-n) t}{N}\right) d t

where we used the Fourier representation for \frac{\sin \pi x}{\pi x},

\displaystyle \int_{-1 / 2}^{1 / 2} e(x t) d t=\frac{\sin \pi x}{\pi x}.

Thus we need to bound

\displaystyle \frac{\pi}{N} \int_{-1 / 2}^{1 / 2} \mathop{\sum\sum}_{m,n\ge 1, m \neq n} \csc(\frac{\pi (m-n)}{N}) a_m e\left( \frac{mt}{N}\right) b_n e\left( \frac{-nt}{N}\right)

Applying the bound N-1 for operator norm with the vectors (a_m e\left( \frac{mt}{N}\right)) and (b_n e\left( \frac{-nt}{N}\right)) we get

\displaystyle \frac{\pi}{N} \int_{-1 / 2}^{1 / 2} (N-1) \|a_n\|_2 \|b_n\|_2 =\left( \pi -\frac{\pi}{N}\right)\|a_n\|_2 \|b_n\|_2

\square

So we see that the operator norm is at most \pi -\frac{\pi}{N}. In fact, it can be show that norm is

\displaystyle \pi-\frac{\pi^{5}} { 2(\log N)^{2}}+O\left(\frac{\log \log N }{(\log N)^{3}}\right)

(Look at Finite sections of some classical inequalities by Wilf). More generally the eigenvalues are given by

\displaystyle \lambda_{n, v}=\pi-\pi^{5} v^{2} / 2(\log N)^{2}+O\left(\log \log N(\log N)^{-3}\right)

K_{\alpha}(m,n) =\frac{1}{m+n+\alpha}

When \alpha=-1, we have

\displaystyle \sum_{n=1}^{\infty} \sum_{m=1}^{\infty} \frac{a_{m} b_{n}}{m+n-1}<\pi|a|_{2}|b|_{2}

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