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Problem 2.10.4 from Kreyszig

Let $X$ and $Y$ be normed spaces and $T_n:X\to Y (n=1, 2,...)$ bounded linear operators. Show that convergence $T_n \to T$ implies that for every $\epsilon>0$ there is an $N$ such that for all $n>N$ and all $x$ in any given closed ball we have $\|T_nx-Tx\|<\epsilon$.

My take on it is below.

$$ \|T_nx-Tx\|=\|(T_n-T)(x)\|\leq \|T_n-T\|\|x\| $$

$x$ is inside the closed ball, let's say that it's a ball with the center $0$ and a radius $r < \infty$: $$ \overline{B}(0,r)=\{x\in X | d(x,0)\leq r\}=\{x\in X |\|x\|\leq r\} $$

Because $T_n\to T$ there exists $N$ such that $\forall n>N:\|T_n-T\|<\epsilon/r$.

Finally, we have $\|T_n-T\|\|x\|\leq \|T_n-T\|r<\epsilon$.

Is the solution correct? Why exactly does the equality $\|T_nx-Tx\|=\|(T_n-T)(x)\|$ hold? How is it possible to prove it if the unit ball isn't centered at $0$?

Thank you for your help.

Konstantin
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2 Answers2

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If $x \in B(y,r)$ then certainly $\|x\|\le M = \|y\| + r$.

Now take $N \in \mathbb{N}$ such that $\|T_n - T\| < \frac{\varepsilon}{M}$.

For $n \geq N$ we have: $$\|T_nx - Tx\| = \|(T_n - T)x\|\le \|T_n - T\|\|x\| < \frac{\varepsilon}{M} \cdot M = \varepsilon$$

mechanodroid
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Your solution is corrert.

The space $L(X,Y)$ of the bounded linear operators $T:X \to Y$ is a linear space.

So $$(T+S)(x)=T(x)+S(x)$$ $$(T-S)(x)=T(x)-S(x)$$

That's why the equality in your statement holds.

Marios Gretsas
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