What is the definition of a bounded operator in an infinite dimensional Hilbert Space?

Question

I am struggling to understand the meaning of a bounded operator in a Hilbert Space.

Does a bounded operator simply means that if it acts on an element of the Hilbert Space, the "result" is bounded?

Answer 1

No.

We say that a linear operator $A: (X, \|\cdot\|_{X}) \to (Y, \|\cdot\|_{Y})$ between two normed vector spaces is bounded if there exists an $L \geq 0$ such that $\|A(x)\|_{Y} \leq L \|x\|_{X}$ for all $x \in X \setminus \lbrace 0 \rbrace$. The infimum of all such $L$, which exists because of the condition $L \geq 0$, is called the norm of the operator $A$ and is denoted by $\|A\|$. One can prove that it is also the minimum of the set of all such $L$'s, i.e. that $\|A(x)\| \leq \|A\|\cdot\|x\|$ for all $x \in X\setminus \lbrace 0 \rbrace$.

It is not too hard to prove that $\|A\| = \sup_{\|x\|=1}\|A(x)\|$, and from there it follows that $A$ is bounded iff it is bounded on the unit sphere $\mathbb{S}_{X} = \lbrace x \in X : \|x\|=1 \rbrace$. If $X$ is finite-dimensional, the unit sphere is compact, and the supremum becomes the maximum. If $X$ is not finite-dimensional, the unit sphere is not compact.

Also, $A$ is contininuous (in the $\varepsilon$-$\delta$ sense) if and only if it is bounded.

Suppose $A$ is continuous: suppose that it's not bounded: that means that $\sup_{\|x\|=1} \|Ax\|= + \infty$, so we can extract a sequence of unit vectors $(x_{n})$ such that $\|Ax_{n}\| \to +\infty$. However, that would mean that the sequence $(y_{n}) = (\frac{x_{n}}{\|Ax_{n}\|})$ satisfies $\|y_{n}\|=\frac{1}{\|Ax_{n}\|} \to 0$, but $\|Ay_{n}\|=1$. That means that $A$ is not continuous at $0$, which is a contradiction with the original assumption.

If we suppose that $A$ is bounded, and if we suppose $x_{n} \to x$, then $\|A(x_{n}) - A(x)\| \leq \|A\|\cdot \|x_{n}-x\|$, so $\|Ax_{n} - Ax\| \to 0$ as well, so $A$ is continuous.

P.S. The proof and the clarity with which I wrote here are probably a bit too speedy to be fully and easily understood by someone reading this for the first time. If continuous mappings between normed vector spaces, Banach spaces, and Hilbert spaces interest you, you can consult any textbook on functional analysis, where all of this is probably written out much more clearly.

Answer 2

First a Point Zero: Nowhere mentioned but seemingly uncontradicted that linear operators are the objects of bounded desire. Adhering to it.
And please accept that a short and basic question happens to eventually receive a longer (hopefully not too lengthy) answer.

The notion(-s) needed to define the property 'bonded' is that of a normed vector space $\big(X,\|\cdot\|\big)$. Each element $\,x\,$ of it has been assigned a finite length, namely $\|x\|$. And for $L(x)$, the image under the linear operator $L:X\to X$, one also has $\|L(x)\|<\infty$.
This is a sort of boundedness, but it does not depend on $L$.
Furthermore, the value $\|L(x)\|$ (if not zero) could be pushed to arbitrary heights by means of scaling the argument vector $x$.

That the linear operator L is bounded means scanning its ratio of stretching $$\frac{\text{Length of image }\|L(x)\|}{\text{Length of argument }\|x\|}$$ while $x$ runs through all of $X$ (well, omit $x=0$) and getting a finite upper limit for it. The ratio is independent of argument scaling, so it's sufficient to evaluate it for every ray in $X$. Thus $x$ may be restricted to only run through the unit sphere of $X$.

If $L$ is bounded, then define the least possible upper limit $$\|L\|_\text{op} := \sup\big\{\|L(x)\|\,\big|\,\|x\|=1\big\}\,\tag{1}$$ and call it the operator norm of L. It's indeed a norm on the vector space comprising all bounded linear operators on $X$, referred to as $\mathscr B(X)$.

From the definition (1) one has the fundamental estimate $$\|L(x)\|\:\leqslant\:\|L\|_\text{op}\cdot\|x\|\quad\forall x\in X\,.$$

Let $T,S\in\mathscr B(X)$, then $\|(T\circ S)(x)\|=\|T(S(x))\|\leqslant\|T\|_\text{op}\cdot\|S(x)\|$ by the preceding. Using again definition (1) it follows that $$\|T\circ S\|_\text{op}\;\leqslant\;\|T\|_\text{op}\cdot\|S\|_\text{op}\,,$$ the operator norm is submultiplicative with respect to composition of linear operators.
That is why $\big(\mathscr B(X),\|\cdot\|_\text{op}\big)$ is a normed algebra of operators.

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Time to approach Hilbert space and to feed the preceding concepts up with four examples.

(1)$\ $ is copied from wikipedia and is about the matrix $\:A=\frac12\left(\begin{smallmatrix}1&3\\-1&3\end{smallmatrix}\right): \mathbb C^2\longrightarrow\mathbb C^2$
to illustrate the ratio of stretching:
It gets maximal for the ray given by the unit vector $\left(\begin{smallmatrix}0\\1\end{smallmatrix}\right)$ $$\left\|\,A\begin{pmatrix}0\\1\end{pmatrix}\right\|_2 \;=\;\left\|\,3\frac{\sqrt{2}}{2}\begin{pmatrix}\sqrt{2}\,/2\\ \sqrt{2}\,/2\end{pmatrix}\right\|_2\;=\;3\frac{\sqrt{2}}{2}\approx 2.12 \;=\;\|A\|_\text{op}$$ It gets minimal for the ray through $\left(\begin{smallmatrix}1\\0\end{smallmatrix}\right)$ $$\left\|\,A\begin{pmatrix}1\\0\end{pmatrix}\right\|_2 \;=\;\left\|\frac{\sqrt{2}}{2}\begin{pmatrix}\sqrt{2}\,/2\\ -\sqrt{2}\,/2\end{pmatrix}\right\|_2\;=\;\frac{\sqrt{2}}{2}\approx 0.707$$ Remark: The maximal and the minimal ratios of stretching are also known as largest and smallest singular value of the operator, respectively.

The $\color{red}{\text{unit sphere in }\mathbb C^2}$ is mapped by $A$ onto a $\color{green}{\text{complex ellipse}}$ whose larger semi-axis equals $\|A\|_\text{op}$.
A real layer looks like

(Graphic from wikipedia)

(2)$\ $ presents a linear operator on an infinite-dimensional pre-Hilbert space which is not bounded:
Consider the Hilbert space $\ell^2(\mathbb N)$ of square-summable sequences and its standard (Hilbert) basis $\{e_n\}_{n=1}^\infty$. Let $\,f$ be the subspace of finite linear combinations of basis vectors, i.e., elements in $\,f$ have only finitely many non-zero entries. $\,f$ is a pre-Hilbert space.

Define $D:f\to f\,$ by $\,e_n\longmapsto ne_n$, then $$\frac{\|De_n\|_2}{\|e_n\|_2}= n$$ which gives a first-hand example of an unbounded set of positive reals. So the linear operator $D$ cannot be bounded.

(3)$\ $ indicates why the $\,\sup\,$ in the definition (1) of the operator norm cannot be replaced by $\,\max$:

Take $M:\ell^2(\mathbb N)\to\ell^2(\mathbb N)$ , defined by $\,M(e_n)= \left(3-\frac1n\right)e_n\,$ for each $n$. As before this is a multiplication operator, but actually defined on the whole of $\ell^2(\mathbb N)$. $$\|M\|_\text{op}\:=\:\sup\big\{3-\tfrac 1n\,\big|\,n\in\mathbb N\big\} \:=\:3\,$$ whereas $3=\|M\|_\text{op}$ as stretch ratio is not realised.
$M\in\mathscr B\left(\ell^2(\mathbb N)\right)$ is a multiple of the identity operator with "some perturbation".

(4)$\ $ Behaviour of a nonlinear operator:
Let $H$ be a Hilbert space. Fix a unit vector $u\in H$, and let $T$ be the operator $T(x)=x+u$.
$T$ translates his argument by a fixed amount and is not linear, neither homogeneous nor additive.

From $\|T(x)\|_2\leqslant\|x\|_2+1\,$ and $\,T(u)=2u\,$ one could conclude $\|T\|_\text{op}=2\,$ after applying definition (1).
Which is misguiding! The ratio of stretching $$\frac{\|x+u\|_2}{\|x\|_2}$$ explodes on each ray in $H$ when $x$ approaches zero.
$\;\Longrightarrow\;$ The property 'bounded' needs revision when leaving the realm of linear operators.

(5)$\ $ proves this answer to be bounded.