Let us consider $R$ with the norm $||x||=|x| $ for the whole discussion. Now the identity mapping $I$ from $R$ to $R$ is unbounded in Calculus but in Functional Analysis treating the same identity mapping $I$ as a linear transformation from the vector space $R$ to the vector space $R$, it becomes bounded. Why have the Mathematicians defined the boundedness in Functional Analysis so much different from that in Calculus?

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    $\begingroup$ The notion of boundedness is the same. We are just using different norms to measure the 'largeness' of a function. For the former, we are using the supremum norm $$ \| T \|_{\sup} = \sup\{ |Tx| : x \in R\} $$ while for the latter, we are using the operator norm $$ \| T \|_{\mathrm{op}} = \sup \{ |Tx| : x \in R \text{ and } |x| = 1 \}. $$ Here, notice that $\| \cdot \|_{\mathrm{op}}$ is only a seminorm if $T : R \to R$ is an arbitrary function, but is indeed a norm on the space of linear operators. $\endgroup$ Mar 28, 2017 at 6:11
  • $\begingroup$ why is the extra condition |x|=1? $\endgroup$
    – Infinite
    Mar 28, 2017 at 6:17
  • $\begingroup$ That is just how the norm is defined. Or you may replace the definition by a more informative one $$ \| T \|'_{\mathrm{op}} = \sup\{ |Tx|/|x| : x \in R \setminus \{0\} \}. $$ (If $T$ is linear, both definitions agree.) You may ask why this definition is useful, and Parish's answer explains this: this norm is designed to capture the continuity of a linear operator. $\endgroup$ Mar 28, 2017 at 6:50
  • $\begingroup$ To expand on Martins' answer, taking the sup norm of a linear operator over an entire space is just $\infty$. This follows from the linearity of the operator: $T(\lambda x) = |\lambda| Tx$. By taking $\lambda$ large, you can make $|T(\lambda x)|$ as large as you want, so $|T|_{\infty} = \infty$. $\endgroup$
    – Seh-kai
    Mar 28, 2017 at 10:25

2 Answers 2


Let $X$ be a Banach space. One way to connect the two definitions of boundedness is to say that a linear map $T : X \to X$ is bounded in the sense of functional analysis, if its restriction to the unit ball $B = \{ x \,:\, |x| < 1 \}$ is bounded in the sense of calculus.

Why the restriction to the unit ball? The only linear map that is bounded in the sense of calculus is $T=0$. Hence for linear maps taking the supremum over the whole space is not an interesting operation. However, taking the supremum over the unit ball (any ball actually) is interesting because of the following property: a linear map $T$ is continuous if and only if $T(B)$ is bounded.

Another interpretation of the name bounded linear map in functional analysis is the following: There is a notion of bounded subsets of Banach spaces (and more generally locally convex spaces), i.e., $B \subseteq X$ is bounded if and only if given any $U \subseteq X$ open with $0 \in U$ there exists $\lambda > 0$ such that $B \subseteq \lambda U$. Then a bounded linear map has the property that it maps bounded sets to bounded sets.


Because we are concerned with linear maps in Functional Analysis and for these linear maps to be continuous with respect to the norm topology it is necessary (infact equivalent) to assume that $||Tx||\leq C||x||$.

Also observe that the words "continuous" and "bounded" are used interchangably in Functional Analysis for this precise reason.

  • $\begingroup$ no I am not asking the continuity, actually due to ONLY such a definition of boundedness Continuity and Boundedness are equivalent in functional analysis. $\endgroup$
    – Infinite
    Mar 28, 2017 at 6:21

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