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Let $𝑓 : 𝑋 → 𝑌$ and $𝑔 : 𝑌 → 𝑍$ be continuous functions between arbitrary vector spaces, and let $ℎ = 𝑔 ◦ 𝑓$ . Suppose $𝑦 = 𝑓 (x)$ and $𝑧 = 𝑔(y) = ℎ(x)$. Give a concrete example where $ \kappa_f (x), \kappa_g (y) \geq 10$ but $𝜅_ℎ (x) \leq 1$.

I am using the relation $\kappa = \frac { ||J(x)||}{||f(x)||/||x||}$ for each of the three functions. I was able to find the upper bound on $𝜅_ℎ(x)$ in terms of $\kappa_𝑓 (x)$ and $\kappa_𝑔 (y)$ (Came out to be $\kappa_ℎ(x) \leq \kappa_f (x)*\kappa_g (y)$). However, now, I am unable to find an example which satisfies the given condition. How can I go about finding an example? Can I generalise anything from this?

Please note that $\kappa$ here is the relative condition number.

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  • $\begingroup$ Just have g be the inverse of f where f is an poorly conditioned but invertible linear map? In this case you need the spaces all the same of course. $\endgroup$
    – Ian
    Mar 2, 2021 at 12:40
  • $\begingroup$ Very similar to math.stackexchange.com/q/4044427/307944 $\endgroup$ Mar 2, 2021 at 15:35
  • $\begingroup$ @CarlChristian Funny, I was unaware and that's my classmate. We're talking about the same question indeed. Thank you for pointing it out. $\endgroup$
    – Iceberry
    Mar 2, 2021 at 15:56
  • $\begingroup$ The assumptions in your stated problem are very weak. A general vector space does not necessarily have a norm. Your formula for the (normwise) relative condition number suggests that your functions can be differentiated, yet this is not assumed in your problem. You should add any extra assumptions as well as the most abstract definition of the condition number used by your textbook. $\endgroup$ Mar 2, 2021 at 16:25

2 Answers 2

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We can take $f(x_1,x_2)=(x_1^n,0)$ and $g(y_1,y_2)=(1,y_2^n)$. Then $h = g \circ f$ satisfies $$ h(x_1, x_2) = g(f(x_1,x_2)) = g(x_1^n,0) = (1,0)$$ As a constant, $h$ will be well-conditioned.

Let us pick the normwise relative condition number and use the infinity norm. In general, if $f : \mathbb{R}^n \rightarrow \mathbb{R}^n$ is differentiable then the condition number can be computed as $$ \kappa_f(x) = \frac{\|Df(x)\|_\infty \|x\|_\infty}{\|f(x)\|_\infty}$$ where $Df(x)$ is the Jacobian of $f$ at the point $x$. It is assumed that $x \not = 0$ and $f(x) \not = 0$. In our case we readily find that $$\kappa_f(x_1,x_2) = \kappa_g(y_1,y_2) = n$$ while $$ \kappa_h(x_1,x_2) = 0.$$


Let me be the first to point out that this reply does not answer the question as it is explicitly stated. Instead, I have strengthened the assumptions to the point where a simple example can be given without losing the surprise that I perceive the professor wanted to give the student.
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  • $\begingroup$ It's a very minor nitpick, but you really want $h$ to map to some nonzero constant instead of actually mapping to zero, so that you don't have a division by zero in $\kappa_h$. But yes, this idea of $g$ just "destroying" what $f$ did will absolutely solve this problem. $\endgroup$
    – Ian
    Mar 2, 2021 at 18:09
  • $\begingroup$ @Ian. Thank you very much for this catch. I have adjusted $h$ accordingly. $\endgroup$ Mar 2, 2021 at 18:16
  • $\begingroup$ @CarlChristian Regarding your example, wouldn't it be $$ \kappa_f(x) = \dfrac{n \max\{|x_1|,|x_2|\}}{|x_1|}?$$ In this case, what you get is $\kappa_f(x) \ge n$. $\endgroup$ Mar 2, 2021 at 18:59
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    $\begingroup$ @PierreCarre You are absolutely correct. I need more sleep :) $\endgroup$ Mar 2, 2021 at 19:21
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I dont' see how that is possible... If you think of the case $X=Y=Z =\mathbb{R}$, you have that $$ \kappa_{g \circ f}(x) = \dfrac{x (g\circ f)'(x)}{(g\circ f)(x)}= \dfrac{x f'(x)g'(f(x))}{g(f(x))} = \dfrac{x f'(x)}{f(x)}\cdot \dfrac{f(x) g'(f(x))}{g(f(x))} = \kappa_f(x)\cdot \kappa_g(f(x)) $$

If $|\kappa_f|, |\kappa_g| \ge 10$, you will fatally have that $|\kappa_{g\circ f}| \ge 100$.

There may be a problem in the original statement of the problem... It is possible to choose particular vector spaces and applications $g,f$ to fulfill the requirement but, for arbitrary vector spaces, it is not always possible to find such applications. Maybe the OP could clarify the intent.

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  • $\begingroup$ In 1D the situation itself is already impossible for linear maps because linear maps all have condition number 1 in 1D. $\endgroup$
    – Ian
    Mar 2, 2021 at 15:56
  • $\begingroup$ @Ian, agreed. But this makes the statement in the OP false, as it speaks about "general vector spaces". $\endgroup$ Mar 2, 2021 at 16:00
  • $\begingroup$ In that regard I agree; it works if "arbitrary" means "you can choose them to suit your needs" but not if it means "they are given" (and the latter is the usual meaning). $\endgroup$
    – Ian
    Mar 2, 2021 at 16:01
  • $\begingroup$ @Ian I find that the language of the original problem is vague. I have added an example that moves us forward. Perhaps my stronger assumptions can be relaxed? $\endgroup$ Mar 2, 2021 at 18:06

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