How to use density argument to obtain inequality? Let $(X, \|\cdot \|)$ and $(Y,\|\cdot\|)$ be a Banach spaces. Assume that $T$ is a linear operator from $X$ to $Y$ ($T:X\to Y$). Assume that  $D\subset X$ is dense in $X.$  The operator $T$ satisfies the inequality $\|Tf\|_{Y} \leq \|f\|_{X}$ for all $f\in D.$ 

Question: Can we expect $\|Tf\|_{Y}\leq \|f\|_{X}$ for all $f\in X$?

Edit: This page tells that: "A common procedure for defining a bounded linear operator between two given Banach spaces is as follows. First, define a linear operator on a dense subset of its domain, such that it is locally bounded. Then, extend the operator by continuity to a continuous linear operator on the whole domain." Can anybody elaborate this or any reference for the details?

The answer to your question is no. There are linear operators $T$ that are bounded on a dense subspace, but not bounded.
See this example which has $Y=\mathbb R$, but is not constructive.
For your edit:
Here, the operator is not given on $X$, but only a dense subset $D$. Therefore we can choose which values $T$ will have on the points $X\setminus D$
You have to do following steps (i will still leave out some details):


*

*define $T f= \lim T f_n$ where $f_n\in D$ such that $f_n\to f$ for all $f\in X$.

*in order to justify 1., show that the limit actually exists (hint: use Cauchy sequences).

*in order to justify 1., show that the definition is independent of the chosen sequence $f_n$.

*show that the resulting operator $T:X\to Y$ is bounded.

