$$T:C_{\mathbb{R}}[0,1] \rightarrow \mathbb{R}$$

Our functional is defined by $T(f) = \int_0^1 f(x) \, dx$. We have to show that this operator is continuous.

I have an answer but it looks really easy, too easy I would say.

We can see that our function $f(x)$ is continuous on the compact set, so it it uniformly continuous and this function is bounded by some constant $M$. So we can write it:

$$|T(f)| = \left |\int_0^1 f(x) \, dx\right | \leq \int_0^1 |f(x)| \, dx \leq \int_0^1 M \, dx = M$$

Our functional is bounded. Hence this implies that functional is continuous, it is the end of the exercise. Am I right?

  • 3
    $\begingroup$ It is really easy but you need to put one more thought in it. You need an estimate $|T(f)|\leq C\|f\|_\infty$ for some $C>0$ and all $f\in C_\mathbb R[0,1]$. Where is your $\|f\|_\infty$-part? $\endgroup$ – Tim B. Jan 20 '17 at 22:36
  • $\begingroup$ My $\left || f \right ||$ can be equal M? Or not? And then I have that $C=1$. $\endgroup$ – FNTE Jan 20 '17 at 22:44
  • $\begingroup$ Yes, absolutely correct, I would probably rather say that $M$ can be chosen as $\|f\|_\infty$ in the above argument but this is what you meant, I guess. $\endgroup$ – Tim B. Jan 20 '17 at 22:44

If you write $\|f\|_\infty$ instead of $M$ (I'm assuming that this is the norm used in the space $C[0,1]$), the proof is much clearer.


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