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Let $f,g: \mathbb R \to \mathbb R$ be continuous. How may I show that the following are also continuous functions

  1. $f+g$
  2. $fg$
  3. $f\circ g$
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3  
what definition of continuous functions do you want to use? –  Alexander Thumm Mar 28 '11 at 15:23

5 Answers 5

up vote 2 down vote accepted

If you use this definition of continuity:

(DEF) $f:X \rightarrow Y$ is continuous at $x_0$ iff $\forall \varepsilon > 0 \exists \delta > 0 : |x - x_0| < \delta \implies |f(x) - f(x_0)| < \varepsilon$

Then for the case (i) you proceed as follows:

you want to show that $f + g$ is continuous at $x_0$. You know, because $f$ and $g$ are continuous that for each $f$ and $g$ respectively you can find a $\delta_f$ and a $\delta_g$ such that $$|x - x_0| < \delta_f \implies |f(x) - f(x_0)| < \frac{\varepsilon}{2} $$ and $$|x - x_0| < \delta_g \implies |g(x) - f(x_0)| < \frac{\varepsilon}{2} $$ for any given $\varepsilon$.

Now you pick $\delta := min(\delta_f, \delta_g)$ and then you use the triangle inequality as follows:

$|(f+g)(x) - (f+g)(x_0)| = |f(x) + g(x) - f(x_0) -g(x_0)| \leq |f(x) - f(x_0)| + |g(x) - g(x_0| < \frac{\varepsilon}{2} + \frac{\varepsilon}{2} = \varepsilon$

where the last inequality follows from how you chose $\delta$.

The cases ii) and iii) are similar.

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2  
at least am getting familiar! i shall learn more and more,though i have limited access to internet –  leopard Mar 30 '11 at 16:28
    
@leopard: that's good! we're always here to help. –  Rudy the Reindeer Mar 30 '11 at 16:53

To go against the current: if you use the definition of continuous from topology,

Let $X$ and $Y$ be topological spaces. A function $f\colon X\to Y$ is continuous if and only if for every open subset $\mathscr{O}\subseteq Y$, $f^{-1}(\mathscr{O})$ is an open subset of $X$.

and you think of $\mathbb{R}$ as having the usual topology (so open sets are arbitrary unions of open intervals), then the proof of (iii) is very easy: let $\mathscr{O}$ be an open subset of $\mathbb{R}$. Then $(f\circ g)^{-1}(\mathscr{O}) = g^{-1}(f^{-1}(\mathscr{O}))$; but $f$ is continuous, so $f^{-1}(\mathscr{O})$ is open, and $g$ is continuous, hence $g^{-1}(f^{-1}(\mathscr{O}))$ is also open. So $f\circ g$ is continuous.

(i) and (ii), though, are a bit more complicated with this definition, though, so you probably want to do $\epsilon$-$\delta$ proofs for those.

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For one and two you can then use the fact that $\mathbb R$ is a topological vector space over $\mathbb R$. =) –  Alexander Thumm Mar 28 '11 at 15:43
3  
@Alexander: Or start by proving that $s(x,y) = x+y$ and $p(x,y)=xy$ are continuous functions $\mathbb{R}^2\to\mathbb{R}$, and then use (iii) to conclude that $s(f,g)$ and $p(f,g)$ are continuous when $f$ and $g$ are continuous. (-; –  Arturo Magidin Mar 28 '11 at 15:45
    
Or we could simply use the hilbert space structure... –  Alexander Thumm Mar 28 '11 at 15:47

"$f+g$":

Let $x\in\mathbb R$ and $\epsilon > 0$.

  1. By continuity of $f$ at $x$ there is $\delta_1>0$ such that for every $y\in\mathbb R$ with $|x-y|<\delta_1$ we have $|f(x)-f(y)|< \epsilon / 2$.

  2. By continuity of $g$ at $x$ there is $\delta_2>0$ such that for every $y\in\mathbb R$ with $|x-y|<\delta_2$ we have $|g(x)-g(y)|< \epsilon / 2$.

  3. Let $\delta = min\{\delta_1, \delta_2\}$. For every $y \in \mathbb R$ with $|x-y|<\delta$ we have $|f(x) + g(x) - f(y) - g(y)| \leq |f(x) - f(y)| + |g(x) - g(y)| < \epsilon/2 + \epsilon/2 = \epsilon$.

Now try the other two as they are similar.

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Use general principles:

(a) The basic operations in ${\mathbb R}$ resp. ${\mathbb C}$ are continuous where defined.

(b) A vector valued function $f=(f_1,f_2)$ is continuous iff the $f_i$ are continuous.

(c) The composition of two continuous functions is continuous.

Now $f\cdot g= p \circ h$ where $h: x\mapsto\bigl(f(x),g(x)\bigr)$ and $p$ denotes the product function.

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@-christian Blatter ; am so interested with the definition f.g –  leopard Mar 30 '11 at 18:49

use the epsilon delta definition

so you want to show f+g is continuous at x. so take a delta for f+g. that delta gives you an epsilon for f and another epsilon for g (by continuity). then you need to come up with an epsilon for f+g.

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"that delta gives you an epsilon" should be "that epsilon gives you a delta", if you would like to use the conventional notation. –  Rudy the Reindeer Mar 28 '11 at 15:37
    
yes, sorry. i switched it around. –  user8799 Mar 28 '11 at 15:39

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