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I have a normed space $E$ and a linear function $f: E \to \mathbb{R}$. Suppose there exists an open set $U$ s.t. $\exists M>0 \; \forall x\in U, \; f(x) < M$. How do I show that $f$ is continuous?

I know that a function is continuous iff for every ball in the codomain centered at $f(x)$, some ball in the domain centered at $x$ is mapped into it.

So pick a point in $\mathbb{R}$, $f(x)$. Since we are dealing with real numbers we know that we can always find a larger number, $a0 \forall x\in U, \; f(x) < M$. That is, the ball $U$ is mapped into the neighborhood of size $M$ centered at $f(x)$. So we have shown that every neighborhood V centered at $f(x) \in \mathbb{R}$ has a ball in the domain $U$ which is mapped into $V$.

Is this correct?

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    No. You showed nothing about about a neighborhood $V$ centered at $f(x)$. You didn't even mention such a neighborhood until your conclusion. You have in no way shown that $U$ maps into every such $V$ (which, btw, it will not, since it is a fixed set). Also, you are given a particular (i.e., fixed) $U$ and $M$. Yet in your proof, you choose an arbitrary value for $M$ and then pretend it has the same property as the fixed one you were told about. You need to pay better attention to the meaning of the phrase "there exists" as opposed to "for every". – 2017-02-15

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To prove this, you need to know the following about open sets in $E$. If you do not already have proofs of these, you will need to prove them:

  • If $Q$ is an open set in $E$ and $e \in E$, then the set $Q + e = \{x + e | x \in Q\}$ is also open.
  • If $Q$ is open and $a \in \Bbb R, a \ne 0$, then $aQ = \{ax | x \in Q\}$ is also open.
  • If $P, Q$ are both open sets, then $P \cap Q$ is also open.

Similar statements are true of open sets in $\Bbb R$:

  • If $O$ is an open set in $\Bbb R$ and $a \in \Bbb R$, then $O + a$ is open, and if $a \ne 0$, then $aO $ is also open.
  • If $N, O$ are opens sets in $\Bbb R$, then $N\cap O$ is also open.

Again, if you have not encountered these facts, you should prove them.

A map is continuous at $x$ if for every open set $O$ containing $f(x)$, there is an open set $Q$ containing $x$ such that $f(Q) = \{f(q) | q \in Q\} \subseteq O$.

With those facts in hand, we can prove continuity of a bounded $f$. We are given that for some non--empty open set $U$ and some $M \in \Bbb R$, for all $x \in U, f(x) \le M$. We work in steps:

  1. Show that we can assume $0 \in U$. That is, there is an open set $U'$ such that $0 \in U'$ and there exists $M' \in \Bbb R$ such that for all $x \in U, f(x) \le M'$.

Hint: Fix some arbitrary $u \in U$, and consider the set $U' - u$.

  1. Show that we can assume $M > 0$ and that $f(x) < M$ for all $x \in U$.

This is trivial, so I won't bother with a hint.

  1. Show that we can also assume for all $x \in U$ that $-M < f(x) < M$ for all $x \in U$.

Hint: Consider $U' = U \cap (-1)U$.

So we have:

  • $U$ is open.
  • $-M < f(x) < M$ for all $x \in U$, with $M > 0$.
  • $0 \in U$.

We can prove continuity in two more steps.

  1. Prove that $f$ is continuous at $0$.

Hint: For $V$ open about $0$ in $\Bbb R$, there is some $\epsilon > 0$ such that the interval $(-\epsilon, \epsilon) \subseteq V$. Find an $a > 0$ such that $f(aU) \subseteq (-\epsilon, \epsilon)$.

  1. Prove that $f$ is continuous at arbitrary $x \in E$.

Hint: If $V$ is an open neighborhood of $f(x)$, then $V - f(x)$ is an open neighborhood of $0$, where we already know that $f$ is continuous. So there is some open neighborhood $Q$ of $0$ in $E$, such that $f(Q) \subseteq V - f(x)$.

Warning: be careful of the notation. It is true that $f(Q) \subseteq V - f(x)$ implies $f(Q) + f(x) \subseteq V$, and that $f(Q + x) = f(Q) + f(x)$, but you can't assume it. You will have to prove that it is so.