Can the Fourier Transform of an L^1 Function be Bounded by its L^1 Norm?

[dRic2]

Hi, I have to show that if $f \in L^1(ℝ^n)$ then:
$$ ||\hat f||_{C^0(ℝ^n)} \le ||f||_{L^1(ℝ^n)}$$
Since $|f(y)e^{-2 \pi i ξ ⋅y}| \le |f(y)|$, using the dominated convergence theorem, it is possible to show that $\hat f \in C^0(ℝ^n)$ but now I don't know how to go on.

Thanks is advance.

 

[mathman]

$\hat{f}(0)=\int f$ should give the result. If $f\ge 0$, then $\hat{f}$ is max at 0. I'm not sure how to finish.

 

[dRic2]

I'm very sorry for the late reply. I don't see why

If f≥0f≥0f\ge 0, then ^ff^\hat{f} is max at 0. I'm not sure how to finish.

.
If that is true I think I solved it, but I don't know how to prove your statement.

PS: In the book I'm reading the author says the result can be achieved without knowing that $\hat f$ goes to $0$ at $\infty$, which he later proves using the result of this proof.

 

[dRic2]

I think I got it! Here is my solution, hope you can tell me if it is correct. (Here I worked in ℝ for the sake of simplicity)

$$|\hat f(ξ)| = \left| \int_ℝ f(y)e^{-2\pi i ξ y}dy \right| \le \int_ℝ |f(y)e^{-2\pi i ξ y}|dy \le \int_ℝ|f(y)|dy = ||f||_{L^1(ℝ)}$$

So basically here I've proved that $|\hat f(ξ)|$ is less than a constant ($||f||_{L^1(ℝ)}$) then, even more so, $\max_{ξ \in ℝ} |\hat f(ξ)| \le ||f||_{L^1(ℝ)}$. This proves the original statement and from here we can also conclude that $\hat f$ is limited.

Am I correct?