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I am using the following definition of continuity:

Let $f: D \subset \mathbb{R^m} \to \mathbb{R^n}$, $x \in D$. Then there exists a function $\delta: \mathbb{R_{>0}} \to \mathbb{R_{>0}}$ so that for all $ \epsilon > 0$ and all $y \in D$ with $\lvert\lvert{x - y}\rvert\rvert < \delta(\epsilon)$ applies $\lvert\lvert f(x) - f(y) \rvert\rvert < \epsilon$.

Now to proof the continuity of the sum of continous functions, I had the following idea:

Let $f: \mathbb{R^m} \to \mathbb{R^n}$ and $g: \mathbb{R^m} \to \mathbb{R^n}$ be continous functions, if $(f+g)$ is countinous, there exists a function $\delta_{f+g}$ so that for all $y\in\mathbb{R^m}$ with $\lvert\lvert x - y\rvert\rvert < \delta_{f+g}(\epsilon')\;\;$ ($\epsilon' > 0, x\in \mathbb{R^m}$) applies $\lvert\lvert f(x)+g(x)-f(y)-g(y)\rvert\rvert < \epsilon'$.
Using triangle inequality: $\lvert\lvert f(x)+g(x)-f(y)-g(y)\rvert\rvert \le \lvert\lvert f(x)-f(y)\rvert\rvert + \lvert\lvert g(x)-g(y)\rvert\rvert < 2\epsilon$.
Now intuitively, I'd say that $\delta_{f+g}$ must be something along $\delta_{f+g}: \epsilon \mapsto \min(\delta_f(\frac{\epsilon}{2}), \delta_g(\frac{\epsilon}{2}))$, but I'm not sure how to proof it.

Another User
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some_user
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1 Answers1

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You have already done all you need you just need to write it down properly.

Let $\varepsilon > 0$ and $\delta_{f+g}(\varepsilon):=\min\{\delta_f\left(\frac{\varepsilon}{2}\right), \delta_g\left(\frac{\varepsilon}{2}\right)\}$. Then for all $x,y\in\mathbb{R}^m$ satisfying $||x-y||<\delta_{f+g}(\varepsilon)$ the inequalities $||f(x)-f(y)||<\frac{\varepsilon}{2}$ and $||g(x)-g(y)||<\frac{\varepsilon}{2}$ are satisfied. Hence, by triangle inequality ...

This way you have shown existence just by defining $\delta_{f+g}$ explicitly and you verify the condition for continuity.

Jfischer
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