# Bounding the ratio of the $l^1$ norms of two vectors to the ratio of their $l^2$ norms

I am having difficulty proving or disproving a claim of the following type: Suppose $\{x_1,\ldots,x_n\}$, $\{y_1,\ldots,y_n\}$ are sets of real numbers satisfying $0 < x_j \leq y_j$ for each $1 \leq j \leq n$, and suppose that $X := \sum_{1 \leq j \leq n} x_j^2$, $Y := \sum_{1\leq j \leq n} y_j^2$. Do there exist absolute constants $c_1,c_2 > 0$ such that \begin{equation*} \frac{\sum_{1 \leq j \leq n} x_j}{\sum_{1 \leq j \leq n} y_j} \leq c_1\left(\frac{X}{Y}\right)^{c_2}. \end{equation*} Evidently, this is interpretable as bounding the ratio of the $l^1$ norms of two vectors by a power of the ratio of their $l^2$ norms.

Obviously, we have a trivial bound with $c_2 = 0$ has $c_1 = 1$, which is undesirable.

A simple application of the Cauchy-Schwarz inequality to the numerator, and of the simple inequality $\left(\sum_{1 \leq j \leq n} y_j^2\right) \leq \left(\sum_{1 \leq j \leq n} y_j\right)^2$ in the denominator gives the bound with $c_2 = \frac{1}{2}$ but $c_1 = n^{\frac{1}{2}}$, which is not suitable, and furthermore, for large $n$ this is worse than the trivial bound mentioned above. \ In my particular application I am taking $n \rightarrow \infty$, my sequence $\{x_n\}_n$ (and thus $\{y_n\}_n$ as well), each term of which depends in some sense on $n$, is growing with $n$, and $\frac{\max_j x_j}{\min_j x_j} \rightarrow \infty$. The same properties are also true of $\{y_j\}_j$. For this reason, reverse Cauchy-Schwarz type inequalities such as that of Polya-Szego do not produce satisfactory improvements when $n$ is significantly larger than either of $X$ or $Y$.

My feeling is that in such generality we ought to allow $c_1$ or $c_2$ to depend on the sequences in question to some degree, but not in a way that reduces to something trivial. For instance, $c_2 = \frac{1}{\log(Y/X)}$ and $c_1 = e$ reduces to the trivial bound by 1, and is thus not interesting.

I welcome any help in this regard, though I would prefer that it be given in the form of a hint, rather than as an explicit proof or counterexample. Thank you in advance.

Update: I was actually able to prove this with $c_1 = 1$ and $c_2 = \frac{1}{4}$. It turns out in my application, however, that I need $c_2 = \frac{1}{2}$. It is clear that $\frac{1}{2}$ is best possible (e.g., by taking $x_1$ and $y_1$ very large compared to each of the remaining $x_j$ and $y_j$, or by allowing zero components and taking $\mathbf{x} := (x,0,\ldots,0)$ and $\mathbf{y} := (y,0,\ldots,0)$; these examples are obviously morally the same). Any absolute $c_1$ is allowable in my application, so any help with the case $c_2 = \frac{1}{2}$ is warmly welcome, even if it does not yield the best constant $c_1$.

• Are your $x_j$ positive, or do you require $|x_j|\le y_j$? – Greg Martin Mar 9 '16 at 22:08
• @GregMartin Yes, thanks. It's fixed in the question now. – Dead-End Mar 9 '16 at 22:13
• The $\l_1/l_2$ norm ratio has been used recently as a "measure" of sparsity for finite sequences, in signal/image processing applications. It saturates at non-sparse flat sequences, and at singletons. Are you open to additional assumptions on your $x$ and $y$? – Laurent Duval Mar 11 '16 at 7:03

## 1 Answer

With $c_2=\frac12$, one cannot take any absolute constant $c_1$ smaller than a constant times $\sqrt n$: an example witnessing this is $x_1=\cdots=x_n=1$ and $y_1=n$, $y_2=\cdots=y_n=1$.

When $c_2=\frac12$, the inequality $$\frac{\sum_{i=1}^n x_i}{\sum_{i=1}^n y_i} \le c_1 \bigg(\frac{\sum_{i=1}^n x_i^2}{\sum_{i=1}^n y_i^2} \bigg)^{1/2}$$ is homogeneous, meaning that it remains unchanged if the $x$ vector or the $y$ vector is scaled by a positive constant. What this means is that the inequalities $x_j\le y_j$ are something of a red herring, since the scaling can reduce any example to one satisfying these inequalities. So Cauchy-Schwarz-type inequalities are probably optimal here. It also means that you're essentially asking for bounds for the ratio $$\frac{\sum_{i=1}^n x_i}{(\sum_{i=1}^n x_i^2)^{1/2}},$$ since the $x$ and $y$ vectors are essentially decoupled.