Does the modulus of a linear operator change continuously with the operator? Let $X,Y$ be real Banach spaces, and let $B(X,Y)$ be the space of bounded linear operators.
Given $T \in B(X,Y)$ the modulus of $T$ is defined to be
$$
\gamma(T):=\inf \{ \,\|Tx\| \, \, | \, \, d(x,\ker T)=1 \}.
$$
It is known that if the image of $T$ is closed in $Y$, then $\gamma(T)>0$.
Suppose that:


*

*$T_n \in B(X,Y)$ is a sequence of operators with closed images, and that $T_n \to T$ in the operator norm. 

*$T$ has a closed image.

*$\dim \ker T_n=\dim \ker T< \infty$. (All the kernels are finite-dimensional and of the same dimension).



Is it true that $\gamma(T_n) \ge c $ for some $ c >0$ independent of $n$?

If the modulus was a continuous map $B(X,Y) \to \mathbb{R}$, then we had $\gamma(T_n) \to \gamma(T)>0$. So the answer would be positive. However, I am not sure the modulus is continuous. Nonetheless, I am interested in the weaker result - must the modulus of a convergent sequence be bounded?
 A: First, observe that in order to prove this statement, it suffices to find a subsequence of $(T_n)$ on which $\gamma(T_n)$ is bounded away from $0$.  If the statement were false, then we could find a subsequence of $(T_n)$ on which $\gamma(T_n)\to 0$, and then applying our argument starting with that subsequence, we would reach a contradiction.
Now let us prove a lemma.

Lemma: Let $(x_n)$ be a bounded sequence in $X$ such that $T_nx_n\to 0$.  Then some subsequence of $(x_n)$ converges to an element of $\ker T$.
Proof: Since $T_n\to T$ uniformly on bounded sets, $Tx_n\to 0$.  Since $\gamma(T)>0$, this implies that $d(x_n,\ker T)\to 0$.  We can thus choose $y_n\in\ker T$ such that $d(x_n,y_n)\to 0$.  Since $(x_n)$ is bounded, so is $(y_n)$.  Since $\ker T$ is finite-dimensional, some subsequence of $(y_n)$ converges to some $y\in \ker T$.  The corresponding subsequence of $(x_n)$ then also converges to $y$.

Now, let $r=\dim \ker T$.  For each $n$, choose a basis $x^1_n,\dots,x^r_n$ of $\ker T_n$ which is "almost orthonormal" in the sense that for each $i$, $x^i_n$ is a unit vector and $d(x^i_n,\operatorname{span}(x^1_n,\dots,x^{i-1}_n))\geq 1/2$.  By the lemma, we may pass to a subsequence and assume that for each $i$, $(x^i_n)$ converges to some $x^i\in\ker T$.  These $x^i$ will also be almost orthonormal, and so in particular will be linearly independent and thus a basis of $\ker T$.
Now suppose that $\gamma(T_n)\to 0$.  This means that we can choose $y_n\in X$ such that $d(y_n,\ker T_n)=1$ for each $n$ but $T_ny_n\to 0$.  Modifying $y_n$ by an element of $\ker T_n$, we may assume that $(y_n)$ is bounded.  By the lemma, we may pass to a subsequence and assume that $(y_n)$ converges to some $y\in\ker T$.  Writing $y$ as a linear combination of $x^1,\dots,x^r$, we see that for large $n$, $y_n$ is close to the corresponding linear combination of $x^1_n,\dots,x^r_n$.  This contradicts the assumption that $d(y_n,\ker T_n)=1$.
Thus $\gamma(T_n)$ does not converge to $0$.  This means we can pass to a subsequence and conclude that $\gamma(T_n)$ is bounded away from $0$, as desired.
