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Definition: Let $f:(M,d)\rightarrow(N,d')$ be a mapping between metric spaces. Then $f$ is continuous at $x\in M$ if for all $\epsilon > 0 $ there exists $\delta$ such that $y\in M$, $d(y,x)<\delta\implies d'(f(y),f(x))\le \epsilon$.

I am given a homework problem where I have to prove a function is continuous. This is the definition I am given. I still struggle in understanding $\epsilon,\delta$ proof. So I want to be sure I understand how to do this before proceeding.

My thought is that if I let $x\in M$ and let $\epsilon > 0$. Then I assume $y\in M$ and $d(y,x)<\delta$. Then show that $d'(f(y),f(x))< \epsilon$ then I have shown that $f$ is continuous. Is this correct?

The statement I am trying to prove is:

Given $[a,b]$, the formula $d(f,g) = \max |f-g|$ defines a metric in the set $C[a,b]$ of continuous functions $f : [a,b] \rightarrow\mathbb{R}$. For $f \in C[a,b]$, let $\Phi(f) = \int_a^b f(x)dx \rightarrow \mathbb{R}$. Show that $\Phi$ is continuous with respect to $d$ and the standard metric $(s, t)\mapsto |s- t|$ in $\mathbb{R}$.

Burgundy
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2 Answers2

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When fleshing out proofs in topology, or any area of mathematics for that matter, I suggest that you simply start writing out what you have in front of you in as concrete form as possible.

Let $f,g \in C[a,b]$ the "distance" between these two in the sup metric is $$ d'(f,g) = \sup_{x \in [a,b]} |f(x) - g(x)|$$ furthermore the distance between their mapping in the respective euclidean metric is $$ d(\Phi(f),\Phi(g)) = \left| \int_a^b f(x) \ dx - \int_a^b g(x) \ dx \right| = \left| \int_a^b (f(x) - g(x)) \ dx\right|$$ Now as you stated if we fix $f \in C[a,b]$ if we can show that $$ \forall \varepsilon > 0 : \exists \delta > 0 : \forall g \in C[a,b] : d'(f,g) < \delta : d(\Phi(f), \Phi(g)) < \varepsilon$$ then this proves that $\Phi$ is continuous at $f$.

The definition of continuity hints at the fact that if we are able to find a suitable relationship between $\varepsilon$ and $\delta$ such that there always exists a suitable distance $\delta$ between the functions $f$ and $g$ then their respective mappings are also close together. So let's get to work!

Fix $f \in C[a,b]$ and lets do some scratch work. The distance between some other function $g \in C[a,b]$ is given above by $$d(\Phi(f), \Phi(g)) = \left| \int_a^b (f(x) - g(x)) \ dx\right|$$ since you are taking topology I assume you are also at least a bit familiar with analysis, in analysis proofs you probably started by trying to find a suitable evaluation in the form of an inequality, we are going to do the same thing here, but we have to keep in mind that the initial space is not the standard euclidean metric, but instead it is the sup metric. We don't have much choice so first we evaluate this integral to a larger one $$ \left| \int_a^b (f(x) - g(x)) \ dx\right| \leq \int_a^b |f(x) - g(x)| \ dx$$ I am going to leave the next step to you, but think closely about what we wish to achieve. We have to somehow get the sup metric in here, and we are extremely close. As a hint consider one more inequality upwards, how could we evaluate this integral in order to get the distance between $f$ and $g$ into the mess? Once you have achieved this, then think about what sort of relationship between $\varepsilon$ and $\delta$ we need. For instance if $$ M \in \mathbb{R} : d'(\Phi(f), \Phi(g)) < M d(f,g) $$ Then if we choose $\delta > 0 : \delta = \frac{\varepsilon}{M}$ then clearly $$d(f,g) < \delta \iff d(f,g) < \frac{\varepsilon}{M}$$ and combining the above inequalities would give us what we want!

Kayle of the Creeks
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The final prove may be written like this:

Let $x\in M$ be arbitrary. Let $\epsilon > 0$ be arbitrary. We take $\delta = ...\text{ some function in } x \text{ and }\epsilon ...$ . Let $y\in M$ with $d(y,x) < \delta$. We have $$d'(f(y),f(x)) < \ldots < \epsilon$$

In the end your prove have to follow the above structure (but might be better formulated ;-) ). The hard thing is to find the $\delta$ for a given $x$ and $\epsilon$ so that you can prove $d'(f(y),f(x)) < \ldots < \epsilon$ under the assumption $d(y,x) < \delta$.

To apply this to your exercise:

  • Fix $f\in C[a,b]$ ( this $f$ will be what $x$ is in the definition ) and $\epsilon > 0$.
  • Take $d'(\Phi(f),\Phi(g))$ for a $g\in C[a,b]$: $$d'(\Phi(f),\Phi(g)) = \left|\int_a^b f(x) dx - \int _a^b g(x)dx\right|$$
  • You are allowed to make the assumption $d(f,g)=\max|f-g|<\delta$ for a $\delta > 0$ which may only depend on $f$ and $\epsilon$.
  • What $\delta$ do you have to pick to prove $d'(\Phi(f),\Phi(g)) < \epsilon$, i.e. $$\left|\int_a^b f(x) dx - \int _a^b g(x)dx\right| < \epsilon$$

Happy proving! ;-)

Stephan Kulla
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