Boundedness of continuous summable function

Question

Let $f\colon\mathbb{R}\to\mathbb{C}$ be a continuous function. If we suppose that $f$ is a $L^1(\mathbb{R;C})$ function too, then can we conclude that $f$ is bounded?

ADD: I asked the preceding question to give an answer for the following problem.

If $f$ and its Fourier Transform are $L^1(\mathbb{R;C})$ functions, then inversion theorem holds almost everywhere. In addition, if $f$ is countinuous and bounded, inversion theorem holds on the whole real line. Does inversion take place when $f$ is just continuous and not bounded?

Answer 1

The word "bounded" can be dropped from the sentence

In addition, if $f$ is continuous and bounded, inversion theorem holds on the whole real line.

Indeed, we are assuming that the Fourier transform of $f$ is integrable. This implies $f$ is in $L^\infty$ and has a continuous representative (as D.F. commented). Assuming "$f$ is continuous" means that we look at this continuous representative for $f$. It is bounded, with $\sup|f| = \|f\|_{L^\infty}$. Indeed, if $|f(a)|>\|f\|_{L^\infty}$, then by continuity there is a neighborhood of $a$ in which $|f |>\|f\|_{L^\infty}$. This contradicts the definition of $\|f\|_{L^\infty}$.

Summary:

If an element of $L^\infty$ has a continuous representative, that representative is bounded.