0

There's already a proof for the following result on Showing that a Volterra operator is well-defined. However I want to ask if my alternative proof is valid.

Exercise: Let $T : C[0,1] \rightarrow C[0,1]$ be defined as $Tf(x) = \int_0^xf(t)dt$. Show that $Tf$ is continuous on $[0,1]$.

Attempt: Take $f \in C[0,1]$. Since $f$ is continuous on a compact interval, then it has an upper-bound $M > 0$ such that $|f(x)| \leq M$ on $[0,1]$. Let $\epsilon > 0$. Make $\delta = \frac{\epsilon}{M} > 0$. Then, if $|x-y| < \delta$, it follows that \begin{align*} |Tf(x)-Tf(y)| & = \left|\int_y^xf(t)dt\right|\\ & \leq \left|\int_y^x|f(t)|dt\right|\\ & \leq M\left|\int_y^xdt\right|\\ & \leq M|x-y|\\ & < M\delta = \epsilon. \end{align*}

user57
  • 689
  • Why do you use ||•|| sometimes but |•| other times? Also, should your $(x - y)$ be $|x-y|$? – Jason Zimba Dec 01 '22 at 14:13
  • @JasonZimba Indeed, I should've used $|\cdot|$ all the time and $(x-y)$ should be $|x-y|$. I've corrected both mistakes. Thank you! – user57 Dec 01 '22 at 14:21
  • @JasonZimba If $M = 0$ I can just pick a new $M* > 0$ and it still is an upper bound for $f$? I am not picking the smaller M but instead, picking some positive M that we know it exists – user57 Dec 01 '22 at 14:42
  • Once you have $|Tf(x)-Tf(y)| \leq M|x-y|$ , it is more usual to apply the Squeeze Theorem. (As here for example: https://math.stackexchange.com/questions/429769/is-an-integral-always-continuous). That makes it unnecessary to define a $\delta$. But your approach looks right to me also. – Jason Zimba Dec 01 '22 at 14:52
  • @JasonZimba Thank you very much! – user57 Dec 01 '22 at 14:53
  • 1
    Should you also include a preliminary observation that $Tf$ exists? (Because a continuous function on a compact interval is Riemann integrable.) – Jason Zimba Dec 01 '22 at 15:07

0 Answers0