A linear bounded operator has to be continuous?

Question

I just found on the wiki https://en.wikipedia.org/wiki/Continuous_linear_operator stating that

An operator between two normed spaces is a bounded linear operator if and only if it is a continuous linear operator.

How to prove that?

Formally, if we define $Z: A \rightarrow B$ is a linear operator between normed spaces $A$ and $B$.

1. How to prove $Z$ is bounded iff $Z$ is continuous?
2. What is the definition of an operator is continuous?

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Some related questions:

If a linear operator between two normed linear spaces is continuous at one point, then it is continuous at all points.

is bounded linear operator necessarily continuous?

Answer 1

Here is a brief outline of the ideas and proofs. Details are everywhere on the internet. Let us fix that our vector spaces are either over $\mathbb R$ or $\mathbb C$.

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Definition: A normed vector space $(V, |\cdot|_V)$ induces a topology, called the norm topology , generated by basis, $B(x,r):= \{y \in V \, : \, |y-x| \le r \}$. This collection forms (exercise, using axioms of norms) a basis. This allows us to talk about conitnuity.

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Norms also induce the idea of boundedness.

Definition: We call a map $f:V \rightarrow W$, between normed vector spaces bounded if exists $c \ge 0$, $$ |f(v)|_W \le c |v|_V.$$ for all $v \in V$.

Definition: An operator between vector spaces $f:V \rightarrow W$ is a linear map.

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So back to your problem. A continuous operator is a continuous linear map between $V,W$ from the induced topology. Precisely, each open ball $B_w(x,r)$ has an open preimage in $V$ under $f$.

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We show that these two are equivalent.

Boundedness =>Continuity. Suppose $f$ is not bounded, (prove)then exists a sequence of points $|x_n | \rightarrow 0$, $|f(x_n)|=1$. If it were continuous, then it is sequentially continuous (prove). So $f(x_n) \rightarrow f(0)=0$. Contradiction.

Continuity=>Boundedness. As $f^{-1}(B_w(0,1))$ is an open set in $V$, then exists an open ball around $0$, $B_v(0,r)$ such that $f(B_v(0,r)) \subseteq B_w(0,1)$.

Exercises: Using linearity show that this is equivalent to boundendness.

Answer 2

For the proof of equivalence, you can find it on Wikipedia.

Since a normed vector space is a metric space with metric induced from the norm, you can just copy the definition of continuity at $x_0$ for real functions of real variable:

$$(\forall\varepsilon >0 )(\exists\delta > 0 )\ \|x-x_0\|<\delta \implies \|Ax-Ax_0\|<\varepsilon.$$

You might be confused by the proof since any function $f$ that satisfies $$\|f(x)-f(y)\|\leq C\| x-y\|$$ for some $C>0$ must be continuous. Try to prove it.

Also, it might be worthwhile for you to try to prove that continuity of linear operator at $0$ implies continuity at all points.