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.
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