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It seems like in a lot of analysis proofs of equalities via inequalities one direction is proven directly, but then the other direction will be shown by inserting, say $-f$ in for $f$ like rudin does in his proof of theorem 6.12 (a) which was typed up here :Theorem 6.12 (a) in Baby Rudin: $\int_a^b \left( f_1 + f_2 \right) d \alpha=\int_a^b f_1 d \alpha + \int_a^b f_2 d \alpha$.

I see why making the substitution above gives us the desired result, but I don't feel entirely comfortable with this method in the sense that I can't really say why it is valid. I fear I may be overthinking it. To be clear, I am not asking why rudin's proof above works, I want to understand why the general trick of substituting in a negative value in for the original is a valid way to prove the reverse inequality.

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    Well, you could have said $f=-g$ for some function $g$, but there's no need to use other letters. If $f$ is any (nice) function so is $-f$ so the result remains true for nice functions. – anon Apr 26 '20 at 05:16
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    It comes down to recognizing when $f$ and $-f$ have the same properties necessary to prove something. Say if $f\geq 0$ and this fact was used in the proof, then $-f$ does not have the property $-f\geq 0$ and this trick could not be used. Keep in mind that Rudin's proof are not the first way someone proved a result, Rudin has just found ways to shorten the proofs yet be clear. – Andrew Shedlock Apr 26 '20 at 05:37

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