Theorem: A linear transformation between two normed spaces is uniformly continuous if it's bounded.

Proof: Let $X_1$ and $X_2$ be normed linear spaces with norms $\| \cdot \|_i,i=1,2$ and let $T$ be a linear transformation between the two spaces. If $T$ is uniformly continuous, then it is continuous at 0 from which it follows that there is a universal $\delta >0$ such that $\|Tx\|_2 \leq 1$ whenever $\|x\|_1 \leq \delta$. Thus, for example, with any $x \neq 0$, we will have $$\|Tx\|_2= \left\|T\frac{\delta x}{\|x\|_1}\right\|_2 \cdot\frac{\|x\|_1}{\delta} \leq \frac{\|x\|_1}{\delta}$$

For the converse, if $x_n \to x$, the fact that $\|T(x-x_n)\|_2 \leq C \|x-x_n\|_1$ means that $T x_n \to Tx$. As C is independent of $x$, the continuity is uniform.

Question: for the converse direction, since we assume it is bounded. Thus, by definition, we have $\|T(x-x_n)\|_2 \leq C\|x-x_n\|_1$ for some finite constant $C$. But how to see that "it means $T x_n \to Tx$"?

  • 2
    $\begingroup$ $x_n \to x$ in $X_1$ and so $\lim_{n \to \infty} \|T(x-x_n)\|_2 \leq C \lim_{n \to \infty} \|x-x_n\|_1 = 0$. This means $\lim_{n \to \infty} Tx_n = Tx$. $\endgroup$
    – Jin
    Feb 11 at 20:46
  • $\begingroup$ @Jin I see. Thanks! $\endgroup$
    – Jonathen
    Feb 12 at 0:46

1 Answer 1


To expand on the answer in the comments, note that if $\|\cdot\|$ is a norm on a vector space $X$, the condition that $\lim_{n\to\infty}y_{n} = y$ means by definition that $\lim_{n\to\infty}\|y_{n} - y\| = 0$. Returning to the question, note that for each $n\in\mathbb{N}$, by the linearity and boundedness of $T$,

$$0 \leq \|T(x_{n}) - T(x)\|_{2} = \|T(x_{n} - x)\|_{2} \leq C\|x_{n} - x\|_{1}.$$

As $\lim_{n\to\infty}0 = 0$ and as $\lim_{n\to\infty}C\|x_{n} - x\|_{1} = 0$, it follows from the squeeze theorem that $\lim_{n\to\infty}\|T(x_{n}) - T(x)\|_{2} = 0$. By definition, $\lim_{n\to\infty}T(x_{n}) = T(x)$.

However, it is unclear how uniform continuity is being shown from this proof. Only sequential continuity has been shown, and it is not clear from what was stated in the proof as to how uniform continuity is obtained from here. However, you can use that the estimate $\|T(x - x_{n})\|_{2} \leq C\|x - x_{n}\|_{1}$ for $n\in\mathbb{N}$ implies Lipschitz continuity with a bit of work.


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