I'm proving the following theorem: Let $\{K_n\}_{n=1}^{\infty}$ be a family of good kernel and $f$ and integrable function on the circle. Then:
$$\lim_{n \to \infty} (f*K_n)(x) = f(x)$$
whenever f is continuous at x.
I wonder if an expression $$|f(x-y) - f(x)| \tag{1}$$ is bounded when $\delta<|y|<\pi$? Otherwise ($|y| < \delta$) I know that $|f(x-y) - f(x)| < \epsilon$ for a fixed $\epsilon$ because $f$ is continuous at x.
If $(1)$ is bounded I would appreciate an explanation.
Asked
Active
Viewed 26 times
0
Hendrra
- 3,066
-
I think the best you can do is $|f(x-y)-f(x)| \le |f(x-y)| + |f(x)| \le 2 M$ where $M$ is a bound on $|f|$. – angryavian Oct 22 '18 at 18:01
-
@angryavian But we only know that $f$ is integrable and continuous at $x$. We don't know anything about $f(x-y)$ – Hendrra Oct 23 '18 at 09:39
-
https://math.stackexchange.com/questions/610054/if-a-function-fx-is-riemann-integrable-on-a-b-is-fx-bounded-on-a – angryavian Oct 23 '18 at 16:38