Consider the Fourier map $\mathcal{F}: L^2(\mathbb{S}^1)\to \ell_2(\mathbb{Z})$. Here $L^2$ denotes $2\pi$-periodic functions on $\mathbb{R}$, and $(\mathcal{F} f)_n = (e_n, f)$, where $(-,-)$ is the inner product on $L^2$ and the $e_n = \frac{1}{\sqrt{2\pi}}e^{inx}$ are orthonormal to each other. It is known that this map is bijective, the proof uses (but is not immediate from) using the inverse map $\ell_2(\mathbb{Z})\to L^2$ that maps $\{c_n\}\mapsto\sum_{n=-\infty}^\infty c_ne_n$. Does this map preserve the inner product as well (and hence is an isomorphism of Hilbert spaces)?
Does the Fourier map preserve the inner product
banach-spacesfourier analysisfourier transformfunctional-analysishilbert-spaces
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Prove the existence of a vector $x$ in the Hilbert space $H$ that satisfies the given inner product.
Note that, by Parseval's inequality, every element of $H$ is written uniquely as $x=\sum_n\langle x,e_n\rangle e_n$. Conversely, given any sequence $(\lambda_n)\in\ell^2$, the vector $x=\sum_n\lambda_ne_n$ is a well-defined element of $H$ and satisfies $\langle x,e_n\rangle=\lambda_n$.
What you are looking for follows directly by the above and the observation that the sequence $(\frac{1}{n})$ is a sequence in $\ell^2$, i.e. $\sum_n\frac{1}{n^2}<\infty$.
A Hilbert space is an inner product space that is complete under the induced norm $\|\cdot\|:=\sqrt{\langle\cdot,\cdot\rangle}$.
Let me give the following counterexample. For $H$ we choose the Sobolev space $H^1(\mathbb{R})$ (sometimes denoted by $W^{1,2}(\mathbb{R})$). This is the subspace of $L^2(\mathbb{R})$ which has weak derivatives that are again in $L^2(\mathbb{R})$. The norm is given by $$ \Vert f \Vert_{H^1(\mathbb{R})} = \sqrt{\Vert f \Vert_{L^2(\mathbb{R}}^2 + \Vert f' \Vert_{L^2(\mathbb{R})}^2},$$ respectively, the scalar product is given by $$ \langle f, g \rangle_{H^1(\mathbb{R})} = \int_\mathbb{R} \overline{f} g + \int_\mathbb{R} \overline{f'} g'. $$ Now consider the inclusion map $$ i: (H^1(\mathbb{R}), \Vert \cdot \Vert_{H^1(\mathbb{R})}) \rightarrow (L^2(\mathbb{R}), \Vert \cdot \Vert_{L^2(\mathbb{R})}, f \mapsto f. $$ The map is clearly injective and it is also bounded as $$ \Vert i(f) \Vert_{L^2(\mathbb{R})} = \Vert f \Vert_{L^2(\mathbb{R})} = \sqrt{\Vert f \Vert_{L^2(\mathbb{R})}} \leq \sqrt{\Vert f \Vert_{L^2(\mathbb{R}}^2 + \Vert f' \Vert_{L^2(\mathbb{R})}^2} = \Vert f \Vert_{H^1(\mathbb{R})}.$$ However, the map is not surjective. For this we need to exhibit an element $L^2(\mathbb{R}) \setminus H^1(\mathbb{R})$. By Morrey's inequality all functions in $H^1(\mathbb{R})$ admit a Hölder continuous representative, thus, $1_{[0;1]}$ would be an example (of course we could check this in a more elementary fashion, but it is good to know Morrey's inequality).
As mentioned in the comments above, the inclusion map is not even an isomorphism of normed spaces on its image. If it was, then there would exist a constant $C>0$ such that for all $f\in H^1(\mathbb{R})$ holds $$ \Vert f \Vert_{H^1(\mathbb{R})} \leq C \Vert f \Vert_{L^2(\mathbb{R})}. $$ For this start with your favourite function $f\in C_c^\infty(\mathbb{R})\setminus \{0\}$ (note that $C_c^\infty(\mathbb{R}) \subseteq H^1(\mathbb{R})$) and for any $\varepsilon>0$ define $f_\varepsilon(x) = \sqrt{\varepsilon} f(\varepsilon x)$. Change of variables tells us $$ \Vert f_\varepsilon \Vert_{L^2(\mathbb{R})} = \Vert f \Vert_{L^2(\mathbb{R})}. $$ However, the derivative picks up an additional factor of $\varepsilon$ and hence, we get $$ \Vert f_\varepsilon \Vert_{H^1(\mathbb{R})}^2 = \Vert f_\varepsilon \Vert_{L^2(\mathbb{R})}^2 + \Vert f_\varepsilon' \Vert_{L^2(\mathbb{R})}^2 = \Vert f \Vert_{L^2(\mathbb{R})}^2 + \varepsilon^2 \Vert f' \Vert_{L^2(\mathbb{R})}^2. $$ Thus, if the inverse of $i$ restricted to its image was continuous, we would have for any $\varepsilon>0$ $$ \Vert f \Vert_{L^2(\mathbb{R})}^2 + \varepsilon^2 \Vert f' \Vert_{L^2(\mathbb{R})}^2 \leq C^2 \Vert f \Vert_{L^2(\mathbb{R})}^2. $$ If $\Vert f' \Vert_{L^2(\mathbb{R})} \neq 0$, then we can send $\varepsilon \rightarrow \infty$ (well, bad naming... if somebody is offended by a large $\varepsilon$, feel free to edit) and get a contradiction. However, $f'$ is continuous, thus, if its $L^2$-norm vanishes, we get that $f'$ vanishes identically and hence, $f$ is constant. This is not possible as $f$ has compact support and is not identically zero.
Best Answer
Just so that this question can be closed, I comment that my question follows from Parseval's Theorem. [Credit to copper.hat and DisintegratingByParts].