can someone help me with this exercise:
Let there f and g be 1-periodic functions both in ${L^2([0,1))}$ . Show that the convolution $$(f*g)(x):=\int_{\mathbb{R}} f(x-y)g(y) \,dy$$ is well defined for all x and that it is 1-periodic and continuous in R.
I have proven that it is well defined and 1-periodic. All i want is some hint as to why it is continuous. Thank you all in advance.
Assuming the integral is meant to be over $[0,1]$, it suffices to note that $$ |(f*g)(x_1) - f*g(x_2)|^2 = \left|\int_{[0,1]} (f(x_1-y) - f(x_2 - y))g(y) \,dy\right|^2\\ \leq \left(\int_{[0,1]} |g(y)|^2 \,dy\right) \left|\int_{[0,1]} |f(x_1-y) - f(x_2 - y)|^2 \,dy\right|, $$ then use the continuity of the shift operator $f(x) \mapsto f(x - a)$ as a linear map over $L^{2}([0,1])$.