Lecture 13: Bounded Inverse Theorem

The Bounded Inverse Theorem

Bounded Inverse Theorem

Let $X$ and $Y$ be Banach spaces, and let $T\colon X\to Y$ be a bounded linear bijection. Then $T^{-1}\colon Y\to X$ is bounded. In particular, there exist constants $A,B>0$ such that $A\|x\|_X\le\|Tx\|_Y\le B\|x\|_X$ for every $x\in X$.

Proof

The upper bound expresses boundedness of $T$. For the lower bound, the Open Mapping Theorem gives continuity of $T^{-1}$. Since $T^{-1}$ is linear, it is bounded. Thus there exists $C>0$ such that

$$ \|T^{-1}y\|_X\le C\|y\|_Y \qquad\text{for every }y\in Y. $$

Substituting $y=Tx$ gives $\|x\|_X\le C\|Tx\|_Y$. Dividing by $C$ gives the lower bound with $A=1/C$.

Example 1: A Finite-Dimensional Linear Map

If $T\colon\mathbb{R}^n\to\mathbb{R}^m$ is linear and surjective, then $m\le n$ by rank–nullity. When $m=n$, surjectivity also gives injectivity, so $T$ is bijective. Every linear map on a finite-dimensional domain is bounded, and every finite-dimensional normed space is complete. Thus the Bounded Inverse Theorem applies for any choice of norms on these spaces.

Application — Which Sequences Are Fourier Coefficients?

Theorem

The Fourier coefficient map $\mathcal{F}\colon L^1(\mathbb{T})\to c_0(\mathbb{Z})$ is not surjective.

Proof

Suppose $\mathcal{F}$ were surjective. By the uniqueness theorem for Fourier coefficients, if $f\in L^1(\mathbb{T})$ and all its Fourier coefficients vanish, then $f=0$ almost everywhere. Thus $\mathcal{F}$ is injective as well. Under our assumption, it would therefore be a bounded linear bijection between the Banach spaces $L^1(\mathbb{T})$ and $c_0(\mathbb{Z})$.

The Bounded Inverse Theorem would then give a constant $A>0$ such that

$$ A\|f\|_{L^1} \le\|\mathcal{F}f\|_\infty =\sup_{n\in\mathbb{Z}}|\widehat f(n)| \qquad\text{for every }f\in L^1(\mathbb{T}). $$

Test this inequality on the Dirichlet kernel $D_N(x)=\sum_{k=-N}^{N}e^{ikx}$. Reading off its Fourier coefficients gives

$$ \widehat{D_N}(n)= \begin{cases} 1, & |n|\le N,\\ 0, & |n|>N. \end{cases} $$

Thus $\|\mathcal{F}D_N\|_\infty=1$ for every $N$. The proposed lower bound would imply

$$ A\|D_N\|_{L^1}\le 1 \qquad\text{for every }N. $$

But we already know that $\|D_N\|_{L^1}\to\infty$. Since $A$ is a fixed positive constant, the inequality fails for sufficiently large $N$. This contradiction proves that some sequences in $c_0(\mathbb{Z})$ are not Fourier coefficient sequences of integrable functions.

Example 2: Two Complete Norms on the Same Vector Space

$$ I\colon (X,\|\cdot\|_1)\to(X,\|\cdot\|_2), \qquad I(x)=x. $$

This map is a linear bijection. The assumed inequality says exactly that $I$ is bounded. By the Bounded Inverse Theorem, $I^{-1}$ is also bounded, so there exists $C>0$ such that $\|x\|_1\le C\|x\|_2$ for every $x\in X$. Taking $A=1/C$ gives

$$ A\|x\|_1\le\|x\|_2\le B\|x\|_1 \qquad\text{for every }x\in X. $$

Hence the two norms are equivalent.

References

SFSU Fourier Analysis by Professor Chun-Kit Lai
Fourier Analysis: An Introduction by Elias M.Stein and Rami Shakarchi