I don't know how to prove this question:

Let $X$ be a compact space, and let $(T_{n})$ be a sequence of positive linear operators on $C(X)$. Also, let $f \in C(X)$ be a strictly positive function. Then if $T_{n}(f) \rightarrow f$ (uniformly), then the sequence $(T_{n})$ is equicontinuous.

Now since this is a strictly positive function, I am aware that the function $f(x) \geq 0$ for all $x \in X$ but how do I goes about in proving the next part of the sentence?

Hope to have someone to shed some light on this. Thank You.

  • $\begingroup$ Isn't it $(T_nf)$ which should be equi-continuous? $\endgroup$ Apr 9, 2012 at 9:42
  • $\begingroup$ In this case, we don't need $f$ to be positive (neither $T$): fix $\varepsilon>0$, so we can find $n_0$ such that $\lVert T_nf-f\rVert_{\infty}\leq\varepsilon$. So the $\delta$ which works for $f$ (which is uniformly continuous) works for $T_nf$, $n\geq n_0$. Then you have only a finite number of uniformly continuous functions. $\endgroup$ Apr 9, 2012 at 9:46
  • $\begingroup$ Hi Davide, based on my understanding of the question, I think I need to show $T_{n}(f) \rightarrow f$ uniformly in order to satisfy the next part of the sentence but I am not so sure. Hope someone can shade some light. $\endgroup$
    – Sandra
    Apr 9, 2012 at 9:47
  • $\begingroup$ Isn't $T_nf\to f$ an hypothesis rather than a thing you have to show? And what would it mean that a sequence of operators is uniformly continuous? $\endgroup$ Apr 9, 2012 at 9:49
  • $\begingroup$ Yes. $T_{n}(f) \rightarrow f$ is the hypothesis. So I am trying to prove that both the hypothesis and the conclusion is true in order to satisfy the sentence. For uniformly continuous, it actually meant $||Tx - Ty|| \leq ||T||||x-y||$. $\endgroup$
    – Sandra
    Apr 9, 2012 at 10:01

1 Answer 1


I'll only deal with real-valued functions, but the complex case is easy to get from this.

A positive operator on $C(X)$ is continuous because its norm satisfies $\|T\| = \|T(1_X)\|_{\infty}$:

Writing $1_X$ for the constant function $x \mapsto 1$, we have for all $g \in C(X)$ that $$ -\|g\|_\infty \cdot 1_X \leq g \leq \|g\|_\infty \cdot 1_X, $$ so positivity of $T$ yields $$ -\|g\|_\infty T(1_X) \leq T(g) \leq \|g\|_\infty T(1_X). $$

and therefore $\|T(g)\|_\infty \leq \|g\|_\infty \cdot \|T(1_X)\|_\infty$. This shows that $\|T\| \leq \|T(1_X)\|_\infty$ and taking $g = 1_X$ we see that a positive operator must have $\|T\| = \|T(1_X)\|_\infty$.

On the other hand, we are given that $\|T_n(f) - f\|_\infty \to 0$, where $f$ is assumed to be strictly positive. By compactness of $X$ and continuity of $f$ there is $x_0 \in X$ such that $0 \lt f(x_0) \leq f(x)$ for all $x \in X$. This tells us that $f(x_0) 1_X \leq f$ and by positivity of the operators $T_n$ we conclude that $0 \leq f(x_0) T_n(1_X) \leq T_n(f)$ whence $\|T_n\| = \|T_n(1_X)\|_\infty \leq \frac{1}{f(x_0)} \|T_n(f)\|_\infty$.

However, since $\|T_nf - f\|_\infty \to 0$, there exists a constant $C$ such that $\|T_n(f)\|_\infty \leq C$ for all $n$. In conclusion, $\|T_n\| \leq \frac{C}{f(x_0)}$ for all $n$, which is equicontinuity of the family of linear operators $T_n$.

  • $\begingroup$ Of course, there are somewhat more direct arguments using that the scalar multiples of $f$ itself dominate all functions, but I think the additional effort is minimal and the information that the operator norm of a positive operator is given by the norm of the image of the constant function $1_X$ is valuable in many contexts. One could also use the uniform boundedness principle, but this looks like overkill. $\endgroup$
    – t.b.
    Apr 9, 2012 at 12:11
  • $\begingroup$ Thanks t.b. this is an interesting argument you had shown here. Lets see if I can think of something else. $\endgroup$
    – Sandra
    Apr 9, 2012 at 12:31
  • $\begingroup$ Oops, I am sorry to let readers know that I had made a mistake in the above question. The proof by t.b. is still correct though. In line 5, the function is $> 0$ and not $\geq 0$. One of such reason is when in line 12 of t.b. proof, the function will be divided by 0 and this is wrong. $\endgroup$
    – Sandra
    Apr 11, 2012 at 4:54
  • $\begingroup$ @Sandra: don't worry: you said strictly positive twice in your post, so it is rather clear $f\gt0$ was intended there. :) $\endgroup$
    – t.b.
    Apr 11, 2012 at 12:58
  • $\begingroup$ @t.b. Thanks I am relieved. After I work on it for some time then I notice my mistake and it gets me worried it might confused the readers, so I immediately make a note on it. Lucky it is not that confusing. $\endgroup$
    – Sandra
    Apr 11, 2012 at 21:12

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