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In Rudin's textbook, "Principles of Mathematical Analysis", theorem 5.3 says:

If $f$ and $g$ are defined on $[a, b]$ and are differentiable at a point $x \in [a,b]$, then $$(f+g)'(x) = f'(x) + g'(x)$$

Rudin said this statement is clear by theorem 4.4, but I tried to prove it by myself. Could you tell me if my way is correct?

My proof:

\begin{align} (f+g)'(x) &= \lim_{t\to x} \frac{(f+g)(t) - (f+g)(x)}{(t-x)}\\ &= \lim _{t\to x} \frac{f(t) + g(t) - f(x) - g(x)}{(t-x)}\\ &= \lim _{t\to x} \frac{f(t)-f(x)}{t-x} + \lim _{t\to x} \frac{g(t)- g(x)}{t-x}\\ &= f'(x) + g'(x) \end{align}

Ayman Hourieh
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InfimumMaximum
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1 Answers1

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It's correct.

I think I'd add a step that says this:

$$ \lim_{t\to x}\left( \frac{f(t)-t(x)}{t-x} + \frac{g(t)-g(x)}{t-x} \right) = \lim_{t\to x} \frac{f(t)-t(x)}{t-x} + \lim_{t\to x}\frac{g(t)-g(x)}{t-x}. $$

The reason I'd want this to be there is:

  • to be explicit about the fact that that is where I relied on the equality between the limit of the sum and the sum of the limits, provided the latter two limits exist;
  • two make it clear at what point the proof relies on the hypothesis that $f$ and $g$ are differentiable at $x$.

You could just replace your second line in your question with the left side of what I wrote above.

Michael Hardy
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