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$$x(t) = E(t) + O(t),$$ where $E(t)$ is an even function and $O(t)$ is an odd function. Prove that $$E(t) = (x(t) + x(-t))/2$$ is unique.

Can I just say that $x(t)$ is unique and $x(-t)$ is unique because both each represent only one particular function, so $E(t)$ must be unique?

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    what do you think is meant by the word unique? – 2017-01-14
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    @WillJagy "Only one" – 2017-01-14
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    Your argument doesn't work (Hint: $1-1=0,3-3=0,\dots$). – 2017-01-14
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    suppose $x=E_1+O_1=E_2+O_2$. then $E_1-E_2=O_2-O_1$. the right-hand side is odd, but the left-hand side is even. the only function that is both even and odd is the zero function. hence $E_1=E_2$. this proves that the decomposition of $x$ into the sum of an even function and an odd function is unique. – 2017-01-14

2 Answers 2

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Hint: $x(t)=E(t)+O(t), x(-t)=E(t)-O(t)$.

And suppose there are two pairs $(E_1,O_1),(E_2,O_2)$ that add up to $x$...

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I think you are misunderstanding the question.

You are not given that $ E (t) = \frac {x (t) + x (-t)}2$.

You must prove:

1) if $E $ and $O $ exist that it must be that $E (t) = \frac {x (t) - x (-t)}2$

2)(trivial) $E $ and $O $ exist.

Proof:

1)$2E (t)=E (t)+O (t)+E (t)-O (t) $

As $E $ is even $E(t)=E (-t)$ and as $O $ is odd $-O (t)=O (-t) $

So $2E (t)=E (t)+O (t)+E (-t)+O (-t)$

$= x (t)+x (-t) $ so

$ E (t) = \frac {x (t) +x (-t)}2$

2) just let $E (t) = \frac {x (t) - x (-t)}2$ and $ O (t) = \frac {x (t) - x (-t)}2$. They clearly satisfy conditions. And by 1) $E $ is the only even function that can. (And so $O (t)=x (t)- E (t) = \frac {x (t) - x (-t)}2$ is the only odd one that does.)