What is the relationship between ratio of vector norms?

Question

If a and b are vectors, then is this $\frac{\Vert a \Vert}{\Vert b \Vert} \le \Vert \frac{a}{b}\Vert$ relation true ? Can it be proved ? ($\Vert \frac{a}{b}\Vert$ is elementwise operation)

Answer 1

Division is not defined for vectors. I suppose by $\frac{a}{b}$ you mean element-by-element division. No relationship can be proved.

from Cauchy–Schwarz Inequality we know that $$\left \| \mathbf{u}.\mathbf{v} \right \|\le\left \| \mathbf{u} \right \|\left \| \mathbf{v} \right \|$$

Put $\mathbf{u}=\mathbf{a}$ and $\mathbf{v}=\frac{\mathbf{1}}{\mathbf{b}}$ (meaning that the elements in $\mathbf{v}$ are the multiplicative inverse of the elements in $\mathbf{b}$):

$$\left \| \frac{\mathbf{a}}{\mathbf{b}} \right \|\le\left \| \mathbf{a} \right \|\left \| \frac{\mathbf{1}}{\mathbf{b}} \right \|$$ where $\mathbf{1}$ is a vector whose elements are all $1$. This is the only thing that we can say and no further conclusion can be drawn regarding $\frac{\left \| \mathbf{a} \right \|}{\left \| \mathbf{b} \right \|}$ (based on the explanations below).

We need to compare $\left \| \frac{\mathbf{1}}{\mathbf{b}} \right \|=\sqrt{\sum_i{\frac{1}{b_i^2}}}$ with $\frac{1}{\left \| {\mathbf{b}} \right \|}=\frac{1}{\sqrt{\sum_i{{b_i^2}}}}$.

From the Jensen's inequality, $\frac{1}{\sum_i{b_i^2}}\le\sum_i{\frac{1}{b_i^2}}$, condidering convexity of $\phi(x)=\frac{1}{x}$.

Hence, $\frac{1}{\left \| {\mathbf{b}} \right \|}\le\left \| \frac{\mathbf{1}}{\mathbf{b}} \right \|$.

Answer 2

The inverse of a vector is not defined, so , if $\vec a$ and $\vec b$ are vectors the writing $\frac{\vec a}{\vec b}$ has no meaning.

If $\vec a$ is a vector and $b$ is an element of the scalar field than

$$ \frac{||\vec a||}{|b|}=||\frac{1}{b}\vec a || $$ by definition of ''norm'' (it is the absolute homogeneity property).