Q:

Convolution of L^p and L^q function is uniformly continuous or not?

This is a homework question (the due date has passed) and I have been thinking of it for a while. We are asked to prove or disprove the following statement:
$f \in L^p(\mathbb{R}), g \in L^p(\mathbb{R}), 1 < p, q < \infty, \frac{1}{p} + \frac{1}{q} = 1$. $f*g(x) = \int f(x-y)g(y)dy$ is uniformly continuous or not?
I know the convolution of a $L^1$ and a $L^{\infty}$ function is uniformly continuous. Below is my attempt, yet I feel I probably did not use Holder's inequality correctly:
Consider a fixed $x \in \mathbb{R}$ and any $a \in \mathbb{R}$, then 
$$|f*g(x-a) - f*g(x)| = | \int f(x-a-y)g(y)dy - \int f(x-y)g(y)dy| \\ 
= | \int f(x-y)g(y-a)dy - \int f(x-y)g(y)dy| \\ 
= | \int f(x-y)[g(y-a) - g(y)]dy| \\ 
\leq \int | f(x-y)[g(y-a) - g(y)] | dy \\ 
\leq ||f||_p ||h||_q
$$
where $h(y) = g(y-a) - g(y), 1 < p, q < \infty, \frac{1}{p} + \frac{1}{q} = 1$. Since the difference between $f*g(x-a)$ and $f*g(x)$ for any fixed $x$ and any $a$ is bounded, I can choose $a$ such that $||h||_q < \frac{\varepsilon}{|| f ||_p}$, so $|f*g(u) - f*g(x)| < \varepsilon$ for any arbitrary $\varepsilon > 0$ and any $u \in (x-a, x+a)$.
I thought I used Holder's inequality correctly, since I was considering some fixed $x$, meaning that I can consider $f(x-y)$ as a function in $y$. Could anyone check my solution? I am really unsure about my use of Holder's inequality. 

A:

The use of Hölder's inequality is fine since $\|f(x-y)\|_p = \|f(y)\|_p$ by translation invariance of the integral. The only thing you need more is that $\|f-f_h\|_q\to 0$ as $h\to 0$ where $f_h(x) = f(x-h)$. This is clear for indicator functions of measurable sets of finite measure, hence for simple functions by linearity and thus for general $L^q$ functions with the usual approximation argument.

