Let f and g be continuous real valued function on $\mathbb{R}^{m}$ prove that f+g are continuous at p

Question

Let f and g be real valued function on $$s \subseteq\mathbb{R}^{m}$$ and if they are both continuous at $\overrightarrow{p}$ then so are f+g.

Let e represent the usual basis for $\mathbb{R}^{m}$. By the linearity of f and g and the Cauchy -Schwarz inequality, for any $\overrightarrow{p} \in\mathbb{R}^{m}$,

$\left| f\left( p\right) +g\left( p\right) \right| ^{2}=\left| p_{1}\left( f\left( e_{1}\right) +g\left( e_{1}\right) \right) +\ldots +p_{m}\left( f\left( e_{m}\right) +g\left( e_{m}\right) \right) \right |^{2} $

$\leq \left( p_{1}^2+\ldots +P_{m}^2\right) \ \left( \left( f\left( e_{1}\right) +g\left( e_{1}\right) \right) ^{2}+\ldots +\left( f\left( e_{m}\right) +g\left( e_{m}\right) \right) ^{2}\right) $

$\left| f\left( p\right) +g\left( p\right) \right| \leq \left\| p\right\| M$

From this it follows that

$\left| f\left( a_{k}\right) -f\left( p\right) +g\left( a_{k}\right) -g\left( p\right) \right| =\left| f\left( a_{k}-p\right) +g\left( a_{k}-p\right) \right| \leq \left\| a_{k}-p\right\| M$

Hence it convergence as $a_{k}$ approaches $p$

Question:

Would this proof be considered correct?

If not what have I done wrong and what would I need to do to get it accepted as being a valid proof.

Answer 1

The following is how I would prove it. Use it as a comparison. Do not take it as the absolute only way to write the proof. In mathematics, we allow personalization in proofs. I am in the camp that a proof requires one to write in complete sentences (where clarity is emphasized over brevity). If you adopt this philosophy, then you may find yourself getting a better response when showing your proof to others (especially professors and teaching assistants!).

Terminology. Let $x_{(\cdot)}$ be a sequence in $\mathbb{R}^n$. Fix $p\in\mathbb{R}^n$. Say $x_{(\cdot)}$ converges to $p$, denoted $x_{(\cdot)}\to p$, if $\lim_{k\to\infty}\Vert p-x_{k}\Vert =0$.

Terminology. Fix $f:U\subseteq\mathbb{R}^n\to\mathbb{R}$, where $U$ is open. Fix $p\in U$. Say $f$ is continuous at $p$ if for each sequence $a_{(\cdot)}$ of elements in $U$, if $a_{(\cdot)}\to p$, then $f(a_{(\cdot)})\to f(p)$.

Proposition. Fix $f,g:U\subseteq\mathbb{R}^n\to\mathbb{R}$, where $U$ is open. Fix $p\in U$. Assume $f$ and $g$ are continuous at $p$. Then $(f+g)$ is continuous at $p$.

Proof. Let $a_{(\cdot)}$ be an arbitrary sequence of elements in $U$. Assume $a_{(\cdot)}\to p$. It remains to show $$(f+g)(a_{(\cdot)})\to (f+g)(p).$$ Observe $$ \begin{align} \Vert(f+g)(p)-(f+g)(a_{(\cdot)})\Vert&=\Vert f(p)-f(a_{(\cdot)})+g(p)-g(a_{(\cdot)})\Vert\\ &\le \Vert f(p)-f(a_{(\cdot)})\Vert+\Vert g(p)-g(a_{(\cdot)})\Vert. \end{align} $$ The right-hand-side goes to zero because $a_{(\cdot)}\to p$ and both $f$ and $g$ are continuous at $p$.

Answer 2

Your proof is correct only for linear functions. However, the functions $f$ and $g$ can be also non-linear. The general proof follows. You want to show that for every $\varepsilon>0$ there is a $\delta-$neighbourhood $P_{\delta}(p)$, where $\delta>0$, so that

$$|(f+g)(p)-(f+g)(x)| \le \varepsilon ~~\text{for}~\text{all}~x\in P_{\delta}(p).$$

You know that from continuity of $f$ and $g$ you get for $\frac{\varepsilon}{2}$ values $\delta_f>0$ and $\delta_g>0$ that

$$|f(p)-f(x)| \le \frac{\varepsilon}{2} ~~\text{for}~\text{all}~x\in P_{\delta_f}(p),$$ $$|g(p)-g(x)| \le \frac{\varepsilon}{2} ~~\text{for}~\text{all}~x\in P_{\delta_g}(p).$$

Hence, you also have for all $x\in P_{\delta}(p)$, where $\delta=\min(\delta_f,\delta_g)$

$$|(f+g)(p)-(f+g)(x)| \le |f(p)-f(x)| + |g(p)-g(x)| \le \frac{\varepsilon}{2} + \frac{\varepsilon}{2} = \varepsilon.$$