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I need to confirm that i am approaching this problem correctly.

Problem: Let $f$ and $g$ be functions $f,g:\mathbb{R}\to \mathbb{R}$

Prove that if $f^2 + g^2$ is bounded then $f$ is bounded.

Attempted Solution:

Definition of bounded: $|f(x)| \leq M$ for all $x \in \mathbb{R}$

Using that definition we can say

$|f^2 + g^2| \leq M$

Then: $|f| = \sqrt{|f^2|} \leq \sqrt{M}$

Therefore we can conclude that $|f|$ is bounded when $|f^2 + g^2|$ is bounded.

Is this true? Or have I gone horrible wrong? Any help is appreciated. Oh and if someone can point me to a good place to learn latex commands that would be sweet, thanks!

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I edited out the (incorrect) tag "set theory". –  Andres Caicedo Feb 16 '11 at 0:41
    
In your first line after "Attempted solution:" you should have for all x, not for all M. The feeling here is that the M is fixed and is a ceiling on the |f^2(x)+g^2(x)|...that the triangle inequality yields something even bigger than this quantity bigger makes things even more crowded... –  Jon Bannon Feb 16 '11 at 0:41
    
@1337holiday: Your argument is incorrect. For instance, how do we know that $|fg|$ is bounded or for that matter how do we even know $|g| < \infty$ and you are allowed to subtract. You need to fill in a lot of gaps in this round about way of proving. –  user17762 Feb 16 '11 at 0:42
    
Your approach is incorrect. All your inequalities are correct, but you haven't finished. All you have is that $|f|$ (a function) is less than or equal to $\sqrt{M+2|fg|}-|g|$ (another function). What you need is that $|f|\le K$ for some constant $K$. One way to fix this would be to show that $\sqrt{M+2|fg|}-|g|$ is itself bounded, because if it is $\le K$ then also $|f|\le K$. The problem is that to prove this seems harder than what is originally asked. (By the way, there is an easier approach.) –  Andres Caicedo Feb 16 '11 at 0:44
    
It would be much easier to remark that $g^2\ge 0$, so you can remove it entirely. For $LaTeX$, there are resources such as web.ift.uib.no/Teori/KURS/WRK/TeX/symALL.html and tutorials on the web. You can also right-click any $LaTeX$ on these pages, choose Show Source, and see how it is done. If you want to use it, put it in dollar signs. –  Ross Millikan Feb 16 '11 at 1:11

1 Answer 1

up vote 8 down vote accepted

There is an easier way, given that squares of real numbers are non-negative, so $f^2 \ge 0$, $g^2 \ge 0$ and $f^2 +g^2 \ge 0$.

If $ f^2 + g^2 \le M$ then $ f^2 \le M$, so $-\sqrt{M} \le f \le \sqrt{M}$.

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Wow thats it? It makes sense and everything but....aghh. Thanks ! –  1337holiday Feb 16 '11 at 1:02

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