# There exists a unique extension $\hat{T}$ of a bounded linear operator $T$.

I am trying to prove the following theorem :

Theorem : Let $$X,Z$$ be Banach (normed) spaces and $$Y$$ be a dense subspace of $$X$$. Let $$T:Y \to Z$$ be a bounded linear operator. Then, there exists a unique extension $$\hat{T}$$ of $$T$$ which is also a bounder linear operator with $$\|\hat{T}\| = \|T\|$$.

Attempted proof :

If $$x \in X$$ then since $$Y$$ is dense over $$X$$, there will exist a sequence $$(x_n)$$ of $$Y$$ with $$x_n \to x$$. This means that for another sequence $$(x_m)$$ of $$Y$$ :

$$\|x_n - x \| \to 0 \implies \|x_n - x_m \| < \varepsilon$$

But since $$T$$ is a bounded linear operator and $$(x_n),(x_m) \in Y$$, it is :

$$\|Tx_n-Tx_m\|\leq \|T\|\cdot\|x_n-x_m\| < \|T\|\cdot \varepsilon$$

which implies that the sequence $$(Tx_n)$$ is Cauchy. But the space $$Z$$ is Banach, so this means that $$Tx_n \to z$$ for some $$z \in Z$$.

Now, if we let $$\hat{T}x \equiv z$$, we beed to prove that $$\hat{T}|_Y = T$$ while $$\hat{T}$$ is linear, bounded and also $$\|\hat{T}\| = \|T\|$$.

Question : I seem to be stuck and out of ideas on how to proceed to proving the final terms needed for my proof. Any hints or elaborations on how to prove the final statements -- $$\hat{T}|_Y = T$$ while $$\hat{T}$$ is linear, bounded and also $$\|\hat{T}\| = \|T\|$$ -- will be very appreciated !

First thing is to show that your extension is well defined. If $$(y_n)$$ is another sequence in $$Y$$ converging to $$x$$ then $$x_n-y_n \to 0$$ so $$\|T(x_n-y_n)\|\leq \|T\| \|x_n-y_n\| \to 0$$. So $$\lim Tx_n=\lim Ty_n$$. Thus $$\hat {T}$$ is well-defined. Now if we start with $$x \in Y$$ we can take $$x_n=x$$ for all $$n$$ so $$\hat {T}x =\lim Tx_n=Tx$$. This proves $$\hat {T}$$ is indeed an extension of $$T$$. Since $$\|\hat {T} x\| =\lim \|Tx_n\| \leq \|T\| \|x_n\|=\|T\|\|x\|$$ we see that $$\|\hat {T}\| \leq \|T\|$$. The definition of norm of an operator immediately gives the reverse in equality (because $$\hat {T}$$ is an extension of $$T$$).