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?

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?


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.


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 $\mathbb C$.

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.

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.

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 $f$.

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.


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