We will first prove that if $B(0,1)=\{b\in B_1:||b||<1\}$, then $T(B(0,1))$ contains an open ball in $B_2$ centered at $0$. Then, since $T$ is linear, we can scale everything properly to show that for any $b_1\in U\subset B_1$ open, there exists $ϵ>0$ such that $B(T(b_1),ϵ)\subset T(U)$

$T$ surjective means everything in $B_2$ gets mapped to by something in $B_1$ ie

So by the Baire Category Theorem, there exists some $n_0\in \mathbb{N}$ such that $\overline{T(B(0,n_0))}$ contains an open ball.

Now, since $T$ is a linear operator, $n_0\overline{T(B(0,1))}$ contains an open ball. ie, $\overline{T(B(0,1))}$ contains an open ball. Equivalently, there exists some $v_0\in B_2$ and $r>0$ such that $B(v_0,4r)\subset \overline{T(B(0,1))}$.

ie, we can pick some $u_1\in B(0,1)$ such that $v_1=T(u_1)\in T(B(0,1))$ such that $||v_0-v_1||<2r$

Then

If $||v||<r$, then

And this means

ie, (in $B_2$), we have $B(0,r)\subset \overline{T(B(0,1))}$. This implies that

for all $n\in \mathbb{N}$

Let $||v||<\frac{r}{2}$. Then $v\in \overline{T\left(B(0,\frac{1}{2})\right)}$ . This implies there exists some $b_1\in B(0,\frac{1}{2})$ such that $||v-T{b}_1||<\frac{r}{4}$

ie, $v-T{b}_1\in \overline{T\left(B(0,\frac{1}{2})\right)}⟹$ there is some $b_2\in B(0,\frac{1}{4})$ such that

continuing in this manner, we obtain a sequence $\{b_k{\}}_{k\in \mathbb{N}}\subset B_1$ such that

1. $||b_k||<2^{-k}$
2. $\left||v-\sum _{k=1}^{n}T{b}_k|\right|<2^{-n-1}r$

The series $\sum _{k=1}^{\mathrm{\infty }}{b}_k$ is absolutely summable (since the norm of the sums are bounded). And since $B_1$ is Banach, this means there exists some $b\in B_1$ such that $b=\sum _{k=1}^{\mathrm{\infty }}{b}_k$

Further, by the triangle inequality, we have

And we can bound this

And since $T$ is a bounded linear operator (and thus continuous), we have

ie, $v\in T(B(0,1))$. Thus $B(0,\frac{r}{2})\subset T(B(0,1))$ in $B_2$.

Suppose $U\subset B_1$ is open and $b_2=T{b}_1\in T(U)$. Then there exists some $ϵ>0$ such that $b_1+B(0,ϵ)=B(b_1,ϵ)\subset U$. And, from above, there exists some $\delta >0$ such that $B(0,\delta )\subset T(B(0,1))$. ie

Since $b_1+B(0,ϵ)$ is contained in $U$. Thus we've found a ball around any $b_2\in T(U)$.

• just need to check all the definitions for the norm
• check that Cauchy sequences in $B_1\times B_2$ consist of Cauchy sequences in each of $B_1$ and $B_2$ (similar to completeness of ${\mathbb{R}}^2$)

Suppose $T\in \mathcal{B}(B_1,B_2)$. Let $\{(u_n,T{u}_n){\}}_n$ be a sequence in $\mathrm{\Gamma }(T)$ such that $u_n\to u$ and $T{u}_n\to v$. Then

since $T$ is continuous. Thus

Define

• ${\pi }_1:\mathrm{\Gamma }(T)\to B_1,{\pi }_1(u,Tu)=u$
• ${\pi }_2:\mathrm{\Gamma }(T)\to B_2,{\pi }_2(u,Tu)=Tu$

Note that $\mathrm{\Gamma }(T)$ is a Banach space since $\mathrm{\Gamma }(T)\subset B_1\times B_2$ is a closed subspace of $B_1\times B_2$ (which is Banach by the corrolary above)

Further, we have ${\pi }_1\in \mathcal{B}(\mathrm{\Gamma }(T),B_1)$ and ${\pi }_2\in \mathcal{B}(\mathrm{\Gamma }(T),B_2)$

(we can see it is bounded by the norm of $B_1\times B_2$ as we defined above, which must also be bounded)

${\pi }_1:\mathrm{\Gamma }(T)\to B_1$ is one-to-one and onto (bijective). Thus $S:={\pi }_1^{-1}:B_1\to \mathrm{\Gamma }(T)$ is also a bounded linear operator

And so $T={\pi }_2\circ S$ is the composition of two bounded linear operators, and is thus itself a bounded linear operator.

ie $T\in \mathcal{B}(B_1,B_2)$