- Given a surjective continuous linear map $f: X \to Y$, image of open unit ball is open.
- Immediate corollary: image of open set is open (translate/scale open unit ball around by linearity) to cover any open set with nbhds.

- Intuition 1: If the map $f$ we bijective, then thm is reasonably believeable given bounded/continuous inverse theorem, since $f^{-1}$ would be continuous, and thus would map open sets to open sets, which would mean that $f$ does the same.
- In more detail: suppose $f^{-1}$ exists and is continuous. Then $f(U) = V$implies $(f^{-1})^{-1}(U) = V$. Since $f^{-1}$ is continuous, the iverse image of an open set ( $U$) is open, and thus $V$ is open.

- Consider the embedding $f : x \mapsto (x, x)$ from $\mathbb R$ to $\mathbb R^2$.
- The full space $\R$ is open in the domain of $f$, but is not open in $\mathbb R^2$, since any epsilon ball around any point in the diagonal $(x, x)$ would leak out of the diagonal.
- Thus, not every continuous linear map maps open sets to open sets.

- Notation: If we have two norms $N$ and $M$ on a space $X$, we denote $B_N(x, r)$ and respectively $B_M$to be the open ball of radius $r$ at point $x$ under norm $N$.
- We will show that if (a) the identity map $(X, N) \to (X, M)$ is continuous, and (b) $B_M(0, 1) \subseteq \overline{B_N(0, r)}^M$, that is, the closure of the ball $B_N(0, r)$ in the $M$ norm,
$B_M(0, 1) \subseteq B_N(0, (1 + \mu)r)$ for any $\mu > 0$.*then* - Inutition: completing under norm $M$ of the open ball on $N$ will continue to trap the ball of $M$at the cost of an infinitesimal radius bump.
- Proof:
- From the hypothesis, we see that every point in $B_M(0, 1)$ is $\epsilon$ close to $B_N(0, r)$ under completion with respect to $M$.
- Thus, we get the inclusion that $B_M(0, 1) \subseteq B_N(0, r) + \alpha B_M(0, 1)$ for
.*any $\alpha > 0$* - We solve the equation $1/(1 - \alpha) = (1 + \mu)$. We can find such a $\beta < 1$, since $\mu > 0$. We pick whatever $\alpha$ we get from this equation to be our $\alpha$.
- We had the equation $B_M(0, 1) \subseteq B_N(0, r) + \alpha B_M(0, 1)$.
- Proof by notation: rewrite as $B_M(0, 1) (1 - \alpha) \subseteq B_N (0, r)$
- This means that $B_M(0, 1) \subseteq 1/(1 - \alpha) B_N(0, r)$, which by definition means that $B_M(0, 1) \subseteq (1 + \mu) (B_N(0, r)$. This gives us what we wanted!
- The geometric picture to keep in mind: $B_M(0, 1)$ can be trapped inside a $B_N(0, r)$ followed by a "small" $B_M(0, 1)$. But by rewriting this "small" $B_M(0, 1)$ itself as a rescaling of $B_N(0, r)$, we can write the full thing as a union of rapidly decreasing balls of the form $B_N(0, r_1), B_N(0, r_2), \dots$.
- All of these will be trapped in $B_N(0, r_1 + r_2 + \dots)$.
- In the above argument, where did we use the continuity of $I$?

- Let $T: X \to Y$ be a one-to-one bounded operator. Then the inverse $T^{-1} : Y \to X$ is also bounded.
- Define a new norm on $Y$ given by $||y||_T \equiv ||T^{-1}(y)||_X$.
- Then this is a norm on $Y$, and we have that $||y||_Y \leq ||T|| ||y||_T$.
- This means that sets measured with $T$ norm are larger than when measured in $Y$ norm.
- This is because $||T(T^{-1}(y)||_Y \leq ||T|| ||T^{-1}(y)||_X$ by the defn of operator norm. Now see that the RHS equals $||T|| ||y_T||$ and we get the desired inequality.
- Thus, the identity map $I : (Y, ||\cdot||_T) \to (Y, ||\cdot||_Y)$ is bounded, since a set that is bounded in the $||\cdot||_T$ norm will be
*smaller*in the $||\cdot||_Y$ norm, and will thus continue to be bouned. - Moreover, since $T$ is continuous, we can check that $(Y, ||\cdot||_T)$ is also a banach space, since a series will converge in $||\cdot||_T$ iff its preimage converges in $X$.
- By the above lemma, this means that the two norms $||\cdot||_Y$ and $||\cdot||_T$ are equivalent.
- So, we get the reverse inclusion, where the norm $||y||_T \leq K ||y||_Y$.
- But that just means that $||T^{-1}(y)|| \leq K ||y||_Y$, or that the operator $T^{-1}$ is bounded!

- Let $T: X \to Y$ be a surjective bounded operator. Then we claim that it's an open map.
- Factorize the map into $T' : X/ket(T) \to Y$ be a bijection, and the projection $\pi : X \to X /ker(T)$will be an open map.
- Let $U \subseteq Y$ be some open set. Since $\pi$ is an open map, we have that $\pi(U)$ is open.
- The latter map is a