Tag: real

Prove that if $ a $ and $ b $ are real numbers with $ a <b < 0$ then $a^2 > b ^ 2 $

Accept $ a $ and $ b $ are real numbers. Prove that if $ a <b <0 $ then $ a ^ 2> b ^ 2 $,

My attempt:

  1. We know that when $ x> y $, then $ -x <-y $,

  2. We also know that when $ x> y ge 0 $, then $ x ^ 2> y ^ 2 $,

Now consider our example:

In the face of that $ a <b <0 $

$$ tag1 a <b $$

$$ tag2 -a> -b $$

$$ tag3 (-a) ^ 2> (-b) ^ 2 $$

$$ tag4 a ^ 2> b ^ 2 $$

Is it right? Suggestions for improvement are welcome.

real analysis – $ int frac {1} {f '(x)} $ diverges

I got this question in a preliminary exam.
$ f $ indicates a positive, continuously differentiable function $ (0, infty) $ so that $ f '> 0 $, If $ f (x) leq Cx ^ 2 $ for some $ C> 0 $ then $ int frac {1} {f '(x)} $ diverges.

There is an indication of the problem with which to determine whether $ int frac {1} {f '(x)} < infty $ then $ int limits_ {a} ^ { infty} frac {1} {f '(x)} to 0 $ as $ a to infty $, I can put the statement in writing in writing $ int f = sum_ {n} int_ {n} ^ {n + 1} f $ and then observe that the above sum converges absolutely and therefore the end of the row (which is the required integral) goes to zero.
I think I should show that in our case now $ int_ {a} ^ { infty} frac {1} {f '(x)} $ does not go to zero as $ a to infty $, But I can not take advantage of that $ f (x) leq Cx ^ 2 $, I do not know if it is useful to notice $ frac {1} {f '(x)} = (f ^ {- 1})' (f (x)) $,

Any help would be appreciated. I would like to have a hint about using this condition.

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real analysis – via quotient space and linear map

I'm reading a book called Functional Analysis written by Peter D Lax

Definition M is a linear map of$$ X rightarrow U $$.$ N_M $ is the set of points that are mapped to zero.
$ R_M $ is the range of M, is the image of X under M in U

He wants me to prove that M maps the quotient space $ X / N_M $ one on one $ R_M $

Here is my confusion first, I think the element of the quotient space is like $ y_1, y_2 $ ,so that $ y_1-y_2 in N_M $So M (y_1-y_2) = 0, then M (y_1) = M (y_n). But that contradicts one on one. Where is my mistake? I do not know how to make mistakes

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