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Prove/disprove: $f+g$ and $f$ are differentiable at $x_0$ $\implies$ $g$ is differentiable at $x_0$

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Suppose $g$ is not differentiable at $x_0$. There are three cases:
Case I: The two following one-sided limits exist, but not equal.$$\lim_{x\to {x_0}^+}\frac{g(x)-g(x_0)}{x-x_0}\neq \lim_{x\to {x_0}^-}\frac{g(x)-g(x_0)}{x-x_0}$$Case II: $g$ is not continuous at $x_0$
Case III: $g$ is undefined at $x_0$

Case I: $$\lim_{x\to {x_0}^-}\frac{(f+g)(x)-(f+g)(x_0)}{x-x_0}=\lim_{x\to {x_0}^-}\Big(\frac{f(x)-f(x_0)}{x-x_0}+\frac{g(x)-g(x_0)}{x-x_0}\Big)=\lim_{x\to {x_0}^-}\frac{f(x)-f(x_0)}{x-x_0}+\lim_{x\to {x_0}^-}\frac{g(x)-g(x_0)}{x-x_0}=L_f+L_{g^-}$$Similarly, with the other one-sided limit:$$\lim_{x\to {x_0}^+}\Big(...\Big)=...=L_f+L_{g^+}$$

Case II: For every type of discontinuity of $g$ we use one-sided limit rules and continuity of $f$ to show that $f+g$ is not continuous and therefore not differentiable.

Case III: That leads to $f+g$ being undefined at $x_0$.

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I haven't yet found a counterexample. If this idea of a proof works, is there a shorter one?

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5 Answers 5

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Instead of proving it by contradiction, you can prove it directly. Note that the following works for the limit approaching $x_0$ from either side: \begin{align*} \lim_{x \rightarrow x_0} \frac{g(x) - g(x_0)}{x - x_0} &= \lim_{x \rightarrow x_0} \frac{(f+g)(x) - (f+g)(x_0)}{x - x_0} - \frac{f(x) - f(x_0)}{x - x_0}\\ &= \lim_{x \rightarrow x_0} \frac{(f+g)(x) - (f+g)(x_0)}{x - x_0} - \lim_{x \rightarrow x_0} \frac{f(x) - f(x_0)}{x - x_0}. \end{align*} The latter two limits exist. You can also use $g = (f+g) - f$ to assure that $g$ is continuous at $x_0$.

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$g=f+g-f$ I think this is enough...

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Hint: First prove that if $a(x)$ is differentiable, then $c \cdot a(x)$ is differentiable for any constant $c$. Next, prove that if $a(x)$ and $b(x)$ are differentiable, then $a(x)+b(x)$ is also differentiable.

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you can prove that if $ h_1$ and $h_2$ are differntiable than $h_1+h_2$ is differentiable and that if $h$ is differentiable than $c\cdot h$ is differentiable (So the set of differentiable function in one point $x_0$ is a subspace of the Vectorial space of the continuos function in $x_0$) In this case $g=(f+g)-f$ and so you have finished

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Rephrasing a bit:

$F(x):= f(x)+g(x)$ ,

$G(x) : (-1)g(x)$ are differentiable at $x=x_0$,

then $F(x) + G(x)$ is differentiable at $x=x_0.$

Used:

1) If $F$ and $G$ are differentiable then $F +G$ is differentiable.

2 If $H$ is differentiable then $-H$ is differentiable

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