Notation: $B_X(a, r) = \{x \in X: \|x - a\| < r\}$.
Proposition. Suppose $X$ and $Y$ are Banach spaces. Let $f: X \rightarrow Y$ be a surjective bounded linear map. Then there exists some $\delta > 0$ such that $\|f(x)\| \geq \delta\|x\|$ for all $x \in X$.
Proof: By the Open Mapping Theorem, $f$ is an open map. Hence we can find some $\delta > 0$ such that $B_Y(0, \delta) \subseteq f(B_X(0, 1))$, so $f(X \backslash B(0, 1)) \subseteq Y \backslash B(0, \delta)$, i.e. $\|f(x)\| \geq \delta$ whenever $\|x\| \geq 1$. Therefore $\|f(x)\| = \|x\|\|f(x/\|x\|)\| \geq \delta\|x\|$.
Here is an application of the above result in the theory of Fourier series.
Theorem. The Fourier transform $\mathcal{F}: L^1(\mathbb{T}) \rightarrow c_0(\mathbb{Z})$ is not surjective.
Proof: Suppose not. By the Proposition, there exists $\delta > 0$ such that $\|\hat{f}\|_\infty \geq \delta\|f\|_1$. But if $\{D_N\}_{N = 0}^\infty$ is the Dirichlet kernel, i.e.
\[ D_N(x) = \sum_{n = -N}^N e^{inx} \]
then we have $\|D_N\|_1 \rightarrow \infty$ and $\|\widehat{D_N}\|_\infty \geq \delta\|D_N\|_1 \rightarrow \infty$, which is absurd.
