# How use density argument to extend operator?

Let $$(X, \|\cdot\|_{X})$$ be Banach space (e.g, $$X=L^p$$) and let $$\mathcal {S}$$ is dense in $$X$$.

Assume that $$T$$ is map such that $$\|Tf\|_{X} \leq \|f\|_{X}\quad \forall f\in \mathcal{S}.$$

My Question is: Can we expect to use density argument to prove that $$\|Tf\|_{X} \leq \|f\|_{X}\quad \forall f\in X?$$ If so, what is natural way to define $$T$$ on $$X$$?

My attempt: Let $$f\in X\setminus \mathcal {S}.$$ Since $$\mathcal{S}$$ is dense in $$X,$$ there exists $$(f_n)\subset X$$ such that $$f_n\to f$$ in $$X.$$ Define $$Tf:= \lim f_n$$. My questions: Can we say the definition of $$Tf$$ is independent of $$f_n$$ (as there might exist $$(g_n)$$ such that $$g_n\to f$$ in $$X$$).

$$\|Tf\|_{X}= \|\lim Tf_n\|_{X} \leq \lim \|Tf_n\|$$ (Can we do this? Can we take limit out side the norm? (I guess norm function is continuous so it make sense? correct?)). And then we get $$\|Tf\|_{X}\leq \lim \|f\|_{X}= \|f\|_{X}.$$

• You can't expect it to be true.
– Rem
Feb 1 at 14:55

To make this more formal and to avoid abuses of notation which at this stage are not yet justified and can also be confusing and/or misleading, let's define a new operator $$\widetilde{T}:X\to X$$ by the following: if $$f\in S$$, then $$\widetilde{T}f = Tf$$ and for $$f\in X\setminus S$$, $$\widetilde{T}^{(f_n)}f = \lim_n T f_n$$ where $$f_n\to f$$ (density of $$S$$ is needed at this stage). Note that $$\widetilde{T}^{(f_n)}f$$ to this point may depend on the choice of $$f_n$$ - we will show it does not. I'm adding the superscript sequence notation just to make the issue clear. You've already thought about it this way, but it helps to write it out as a new operator (which it technically is).
We'll first show that the operator norm condition extends to $$\widetilde{T}$$. As you note, $$\|\widetilde{T}^{(f_n)}f\| = \|\lim_n Tf_n\| = \lim_n \|Tf_n\| \le \lim_n \|f_n\|$$ as the norm is continuous (almost by definition since limits in $$X$$ come from the norm). Since $$f_n\to f$$ (meaning $$\lim_n \|f_n-f\| = 0$$), $$\lim_n \|f_n\| = \|f\|$$ and so $$\|\widetilde{T}^{(f_n)}f\| \le \|f\|$$.
Let us show that $$\widetilde{T}$$ is well-defined, i.e. if there exists another sequence $$g_m\to f$$, then $$\widetilde{T}^{(f_n)}f = \lim_n Tf_n = \lim_m Tg_m = \widetilde{T}^{(g_m)} f$$ as well. The sketch below is the basis for the argument which I will leave you to fully formalize:
\begin{align} \|Tf_n - Tg_m\| &= \|T(f_n - g_m)\| \\ & \le \|f_n - g_m\| \\ &= \|f_n - f + f - g_m\| \\ & \le \|f_n - f\| + \|f - g_m\| \\ & \to 0 \end{align}
Thus as $$n$$ and $$m$$ tend to infinity, $$Tf_n$$ and $$Tg_m$$ converge to the same element since limits are unique in Banach spaces. Therefore we can drop the $$\widetilde{T}^{(f_n)}$$ notation in favor of a singular $$\widetilde{T}$$ notation.