Define the Fourier Transform as
$$
\hat{f}(\xi)=\int_{\mathbb{R}^n}f(x)e^{-2\pi ix\cdot\xi}\,\mathrm{d}x\tag{1}
$$
In this answer, it is shown that in $\mathbb{R}^n$
$$
\|\xi\hat{f}\|_2\|xf\|_2\ge\frac{n}{4\pi}\|\hat{f}\|_2\|f\|_2\tag{2}
$$
If we define $\tau_yf(x)=f(x+y)$, then
$$
\begin{align}
\widehat{\tau_yf}(\xi)
&=\int_{\mathbb{R}^n}f(x+y)e^{-2\pi ix\cdot\xi}\,\mathrm{d}x\\
&=e^{2\pi iy\cdot\xi}\int_{\mathbb{R}^n}f(x)e^{-2\pi ix\cdot\xi}\,\mathrm{d}x\\
&=e^{2\pi iy\cdot\xi}\hat{f}(\xi)\tag{3}
\end{align}
$$
Then The Plancherel Theorem and Fourier Inversion give
$$
\begin{align}
\|(\xi-b)\hat{f}\|_2\|(x-a)f\|_2
&=\|\xi\tau_b\hat{f}\|_2\|x\tau_af\|_2\\
&=\|\xi e^{-2\pi ia\cdot\xi}\tau_b\widehat{\tau_af}\|_2\|x\tau_a f\|_2\\
&=\|\xi\tau_b\widehat{\tau_af}\|_2\|x\widehat{\widehat{\tau_a f}}\|_2\\
&=\|\xi\tau_b\widehat{\tau_af}\|_2\|xe^{-2\pi ib\cdot x}\widehat{\tau_b\widehat{\tau_a f}}\|_2\\
&=\|\xi\tau_b\widehat{\tau_af}\|_2\|x\widehat{\tau_b\widehat{\tau_a f}}\|_2\\
&\ge\frac{n}{4\pi}\|\tau_b\widehat{\tau_af}\|_2\|\widehat{\tau_b\widehat{\tau_a f}}\|_2\\
&=\frac{n}{4\pi}\|\hat{f}\|_2\|f\|_2\tag{4}
\end{align}
$$
In case $\int_\mathbb{R} \vert x \vert^2 \vert f(x) \vert^2 dx =\infty$, then the inequality is trivial. Thus, we assume for the rest $\int_\mathbb{R} \vert x \vert^2 \vert f(x) \vert^2 dx <\infty$.
By the fundamental theorem of calculus we get
$$ \int_\mathbb{R} \vert f (x) \vert^2 dx = \int_\mathbb{R} \overline{f}(x) x \frac{1}{x} \left( f(x_0) + \int_{x_0}^x f'(y) dy \right)dx. $$
First we consider the second term. We have
$$ \int_\mathbb{R} \overline{f}(x) x \frac{1}{x} \int_{x_0}^x f'(y) dy dx
\leq \left( \int_\mathbb{R} x^2 \vert f(x) \vert^2 dx\right)^{1/2}\left( \int_\mathbb{R} \frac{1}{\vert x \vert} \int_{x_0}^x \vert f'(y) \vert^2 \right)^{1/2}
\leq \left( \int_\mathbb{R} x^2 \vert f(x) \vert^2 dx\right)^{1/2}\left( 4\int_\mathbb{R} \vert f'(x) \vert^2 dx \right)^{1/2}, $$
where we used Hardy's inequality to pass to the last line.
For the second term we note that we just need to show that $f\in L^1(\mathbb{R})$, then we just choose $x_0$ such that $f(x_0)$ gets smaller and smaller to eliminate this term. For this we use Cauchy-Schwarz and that $f$ is locally bounded
$$ \int_\mathbb{R} \vert f(x) \vert dx = \int_{[-1, 1]} \vert f(x) \vert dx + \int_{\mathbb{R}} \vert x \vert \vert f(x) \vert \frac{1}{x} dx
\leq 2 \sup_{x\in [-1,1]} \vert f(x) \vert + \left( \int_\mathbb{R} x^2 \vert f(x) \vert^2 \right)^{1/2} \left( 2 \int_1^\infty \frac{1}{x^2} dx \right)^{1/2} <\infty. $$
Hence, we have shown so far
$$ \left( \int_\mathbb{R} \vert f (x) \vert^2 dx \right)^2
\leq 4 \left( \int_\mathbb{R} x^2 \vert f(x) \vert^2 dx\right)\left( \int_\mathbb{R} \vert f'(x) \vert^2 dx \right). $$
Finally we note that by Parseval's identity we have
$$ \int_\mathbb{R} \vert f'(x) \vert^2 dx = \int_{\mathbb{R}} \vert \widehat{f'}(\xi) \vert^2 d\xi. $$
Now use that $\widehat{f'}(\xi)= 2\pi i \xi \widehat{f}(\xi)$ to conclude.
Best Answer
Let $1 \leqslant p \leqslant +\infty$. For $x\in \mathbb{R}$ [it works with the same argument in $L^p(\mathbb{R}^n)$ for any $n > 0$], let $\tau_x \colon L^p(\mathbb{R}) \to L^p(\mathbb{R})$ be given by
$$\tau_x(f) \colon u \mapsto f(u-x).$$
By the translation-invariance of the Lebesgue measure, $\tau_x$ is an isometry of $L^p(\mathbb{R})$ for every $x$.
Further, let $\rho \colon L^p(\mathbb{R}) \to L^p(\mathbb{R})$ be the reflection, $\rho(f) \colon u \mapsto f(-u)$. Then $\rho$ is also an isometry of $L^p(\mathbb{R})$ since the Lebesgue measure is also invariant under reflection.
Thus for any $f,g\in L^2(\mathbb{R})$ we can write
$$(f\ast g)(x) = \int_\mathbb{R} f(x-u)g(u)\,du = \int_\mathbb{R} \bigl(\rho(\tau_x(f))\bigr)(u)g(u)\,du.$$
For real-valued functions, we then see
$$(f\ast g)(x) = \langle \rho(\tau_x(f)), g\rangle_{L^2},\tag{1}$$
and for complex-valued functions, we need to conjugate one of the arguments of the inner product, and then note that complex conjugation is also an isometry of $L^p(\mathbb{R}^n)$ for all $1\leqslant p \leqslant +\infty$.
Now we can apply the Cauchy-Schwarz inequality to the right hand side of $(1)$ to obtain
$$\lvert (f\ast g)(x)\rvert \leqslant \lVert \rho(\tau_x(f))\rVert_2\cdot \lVert g\rVert_2\tag{2}.$$
But since $\rho$ and $\tau_x$ are isometries, the right hand side of $(2)$ is independent of $x$, and we have
$$\lvert (f\ast g)(x)\rvert \leqslant \lVert f\rVert_2\cdot\lVert g\rVert_2.\tag{3}$$
By $(3)$, the convolution $f\ast g$ is uniformly bounded, and taking the supremum over all $x\in\mathbb{R}$ yields
$$\lVert f\ast g\rVert_\infty = \sup_{x\in\mathbb{R}} \lvert (f\ast g)(x)\rvert \leqslant \lVert f\rVert_2\cdot\lVert g\rVert_2.\tag{4}$$
Since $f\ast g$ is continuous and $(f\ast g)(x) \to 0$ for $\lvert x\rvert \to \infty$ - note that these properties need not be assumed, they can be proved under the assumptions - the supremum in $(4)$ is actually attained, and can be replaced by $\max$ if one so wishes.
Basically the same argument, using the more general Hölder inequality instead of the Cauchy-Schwarz inequality shows that for $1\leqslant p,q\leqslant +\infty$ with $\frac{1}{p} + \frac{1}{q} = 1$ we have
$$\lVert f\ast g\rVert_\infty \leqslant \lVert f\rVert_p\cdot \lVert g\rVert_q$$
for all $f\in L^p(\mathbb{R})$ and $g\in L^q(\mathbb{R})$.
For $1\leqslant p < +\infty$, the continuous functions with compact support are dense in $L^p(\mathbb{R})$, and from that follows that for any fixed $f\in L^p(\mathbb{R})$ the function $x \mapsto \tau_x(f)$ is continuous, which shows that $f\ast g$ is continuous - if $p = \infty$, then $q = 1 < \infty$, and we note that we have
$$(f\ast g)(x) = \int_\mathbb{R} f(u)g(x-u)\,du,$$
so the continuity of $x\mapsto \tau_x(g)$ yields the continuity of $f\ast g$.
For $1 < p,q < +\infty$, we have
$$\lim_{\lvert x\rvert\to\infty} (f\ast g)(x) = 0,$$
as one can again see by using the denseness of $C_c(\mathbb{R})$ in $L^p(\mathbb{R})$ and $L^q(\mathbb{R})$ then, but that need not hold in the $\{ p,q\} = \{1,\infty\}$ case.