4
$\begingroup$

It is known that if a function $f:\Bbb R\to \Bbb R$ is continuous then its graph is closed.

Proof. Let $(x_n)_{n\in\Bbb N}$ be a sequence in $\Bbb R$ so that the sequence $(x_n,f(x_n))_{n\in\Bbb N}$ is convergent in $\Bbb R^2$ at a point $(x,y)\in\Bbb R^2$. Then $$(x_n,f(x_n)) \xrightarrow{n\to \infty} (x,y) \Longrightarrow x_n\to x \ \ \& \ \ f(x_n)\to y $$ Now, from the continuity of $f$ we have that

$$x_n \to x \Longrightarrow f(x_n)\to f(x)$$

and from the uniquence of the limits we assume that $y=f(x)$.

So, $\lim_{n\to\infty}(x_n,f(x_n))=(x,f(x))\in G(f)$ and so $G(f)$ is closed.


We know that the converse is not true in general, to wit if a real function $f$ has closed graph $G(f)=\{(x,f(x))|x \in\Bbb R \}\subset \Bbb R^2$ we cannot assume that $f$ is continuous. One counter example is the function: $$f: \Bbb R \to \Bbb R, \ \ f(x)=\begin{cases} \text{$\frac{1}{x} \ \ \ \ $ if } x \neq 0 \\ \text{$0 \ \ \ \ \ $ if } x= 0 \end{cases}$$

since $f$ is discontinuous at $x=0$ and $G(f)=\{ (0,0) \}\cup \{(x,\frac{1}{x})|x\in\Bbb R\setminus \{0\}\}$ is closed because both of the sets are closed.

but if we add that $f$ is bounded then it can be proved that $f$ is continuous. I am having problem in the proof. Here is my attempt:


Attempt of a proof. Let $(x_n)_{n\in\Bbb N}$ be a real sequence that converges to some $x\in\Bbb R$. We need to prove that $f(x_n)\xrightarrow{n\to \infty}f(x)$. We have that $(x_n,f(x_n))\in G(f) \ \ \forall n\in\Bbb N$ and that $((x_n,f(x_n))_{n\in\Bbb N}$ is bounded on $\Bbb R^2$ (since $(x_n)_{n\in\Bbb N}$ is convergent and $f$ is bounded). So from the Bolzano-Weierstrass theorem there exists a subsequence $(x_{k_n})_{n\in\Bbb N}$ of $(x_{n})_{n\in\Bbb N}$ so that $(x_{k_n},f(x_{k_n}))_{n\in\Bbb N}$ converges. Now, because $G(f)$ is a closed set $\exists x'\in\Bbb R :(x_{k_n},f(x_{k_n}))\to (x',f(x')) $ and so $x_{k_n}\to x'$ and $f(x_{k_n})\to f(x')$. Moreover, $x_n\to x \ \ \& \ \ x_{k_n}\to x' \Longrightarrow x=x'$ and so $f(x_{k_n})\to f(x)$.

I cannot go any further than this. If I could prove that $x_n\to x \ \ \& \ \ G(f)$ is closed $\ \& \ \ f $ is bounded $\Longrightarrow (f(x_n))_{n\in\Bbb N}$ converges, then I could end the proof.

$\endgroup$
  • $\begingroup$ If the main point of your question is to get some input on your own proof - verifying it, suggesting improvements, finding mistakes - as opposed to asking other users for any proof of the claim in question - then you should indicate that by using (proof-verification) tag. See the tag-info for more details. $\endgroup$ – Martin Sleziak Dec 9 '16 at 15:18
2
$\begingroup$

Very good so far. To finish it, assume that $\bigl(f(x_n)\bigr)$ didn't converge to $f(x)$. Then there would be an $\varepsilon > 0$ and a subsequence $(x_{n_k})$ of $(x_n)$ with $\lvert f(x_{n_k}) - f(x)\rvert \geqslant \varepsilon$ for all $k$. Apply your argument to the sequence $(y_k)$ given by $y_k = x_{n_k}$ to obtain the contradiction that $f(y_{k_m}) \to f(x)$ for some subsequence of $(y_k)$, although by construction $\lvert f(y_k) - f(x)\rvert \geqslant \varepsilon$ for all $k$. So the initial assumption is untenable, and we conclude that $f(x_n) \to f(x)$. Since the sequence $(x_n)$ converging to $x$ was arbitrary, $f$ is continuous at $x$. Since $x$ was arbitrary, $f$ is globally continuous.

$\endgroup$
3
$\begingroup$

Another way to think about this (not following the OP's approach, but using compactness in a slightly different way):

To show that $f$ is continuous, it's enough to show that the restriction $f_{|I}: I \to \mathbb R$ is continuous for each closed interval $I \subset \mathbb R$; this is what we'll do.

Let $\Gamma_f \subset \mathbb R^2$ denote the graph of $f$, which is closed by assumption. Then $\Gamma_{f_{|I}} := \Gamma_f \bigcap (I \times \mathbb R)$, which is the graph of $f_{|I}$, is closed in $I\times \mathbb R$. The assumption that $f$ is bounded shows that it is also bounded, and so it is a compact subset of $I\times \mathbb R$.

Now it is a general property of graphs that the projection $\Gamma_{f_{|I}} \to I$ given by $(x,y) \mapsto x$ is continuous and bijective. A continuous bijection between compact topological spaces is necessarily a homeomorphism. Taking the inverse homeomorphism we obtain a homeomorphism $I \to \Gamma_{f_{|I}}$, which when composed with the (continuous) projection $\Gamma_{f_{|I}} \to \mathbb R$ defined by $(x,y) \mapsto y$, gives the function $f_{|I}: I \to \mathbb R$, but now realized as a composite of continuous functions. This shows that $f$ is continuous.

(We also see that we don't need to assume that $f$ is bounded, but just that $f$ is bounded when restricted to bounded subsets of $\mathbb R$ --- a property that every continuous function on $\mathbb R$ satisfies; of course, this is also clear with the OP's proof strategy (as completed in the other answers). )

$\endgroup$
2
$\begingroup$

Here's a summary of the ideas presented above.

If $\lim_{n\to\infty} x_n =x$, yet $f(x_n)$ does not converge to $f(x)$, WLOG, $\limsup_{n\to\infty} f(x_n) > f(x)$ (otherwise replace by $-f$), then there exists a subsequence $(x_{n_k})$ such that $\lim_{k\to\infty} f(x_{n_k}) = \limsup_{n\to\infty} f(x_n)> f(x)$. Since $G(f)$ is closed, $(x,\limsup_{n\to\infty} f(x_n))\in G(f)$, in contradiction to $G$ being the graph of a function.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.