The answer is "Yes" though the construction (albeit very standard) is slightly non-obvious.

First, we'll notice that for $f\in L^2$, you can make $f$ and $\hat f$ anything you want on $I=[-0.1,0.1]$, say. That is just because the product $PQ$ of the projections $Pf=f\chi_I$ and $Qf=(\hat f\chi_I)^\vee$ is a contraction in $L^2$, so to solve $Pf=Pg, Qf=Qh$, you just put $f=(g+h)-(Ph+Qg)+(QPh+PQg)-(PQPh+QPQg)+\dots$. Notice that if $g,h$ are real even, the series will consist of real even functions as well, so $f$ will be real even too.

Now just choose some $C^\infty$ even real mollifier $\psi$ with support $(-0.01,0.01)$, $\int_{\mathbb R}\psi=1$, and positive $\hat\psi$. Solve the problem $f|I=0, \hat f|I=\frac{1}{\hat\psi}$ in $L^2$.

Take $F=f*\psi$. It will remain $0$ on $J=\frac 12I$ but we shall get $\hat F_I=\hat f\hat\psi|I=1$. and $\hat F$ will decay fast. Now convolve with $\psi$ on the Fourier side. You'll get a function $G\in S$ such that $G$ is real even, $G=0$ on $J$ and $\hat G=1$ on $J$. In particular, $$ \int_{\mathbb R}G(t)=1,\quad \int_{\mathbb R}G(t)t^{2k}=0\text{ for }k\ge 1 $$

Since $G$ is even, we can say the same for the integrals from $0$ to $+\infty$, with $2G$ instead of $G$. It remains to change the variable, i.e., to take $$ 2G(t)=2t H(t^2) $$ on $(0,+\infty)$ (since $G$ vanishes near the origin, that will result in $H\in S$ still). The function $H$ has exactly the properties you wanted.

Edit: By Alexei's request, here is the proof that $PQ$ is contracting. We have $\|PQ\|^2=\|PQP\|$. The operator $PQP$ has the kernel $\chi_I(x)\chi_I(x')\int_Ie^{2\pi i(x-x')y}\,dy$, which is supported on $I\times I$ and is bounded by $I$ in absolute value. Thus, by the HS bound, or by the Schur test, we have $\|PQP\|\le |I|^2$, i.e., $\|PQ\|\le |I|$.

It is worth noting that it is actually contracting for an interval $I$ of arbitrary length or even for a set $I$ of finite Lebesgue measure, but the proofs of those facts are not so simple.

The answer is "Yes" though the construction (albeit very standard) is slightly non-obvious.

First, we'll notice that for $f\in L^2$, you can make $f$ and $\hat f$ anything you want on $I=[-0.1,0.1]$, say. That is just because the product $PQ$ of the projections $Pf=f\chi_I$ and $Qf=(\hat f\chi_I)^\vee$ is a contraction in $L^2$, so to solve $Pf=Pg, Qf=Qh$, you just put $f=(g+h)-(Ph+Qg)+(QPh+PQg)-(PQPh+QPQg)+\dots$. Notice that if $g,h$ are real even, the series will consist of real even functions as well, so $f$ will be real even too.

Now just choose some $C^\infty$ even real mollifier $\psi$ with support $(-0.01,0.01)$, $\int_{\mathbb R}\psi=1$, and positive $\hat\psi$. Solve the problem $f|I=0, \hat f|I=\frac{1}{\hat\psi}$ in $L^2$.

Take $F=f*\psi$. It will remain $0$ on $J=\frac 12I$ but we shall get $\hat F_I=\hat f\hat\psi|I=1$. and $\hat F$ will decay fast. Now convolve with $\psi$ on the Fourier side. You'll get a function $G\in S$ such that $G$ is real even, $G=0$ on $J$ and $\hat G=1$ on $J$. In particular, $$ \int_{\mathbb R} G(t) \, dt = 1,\quad \int_{\mathbb R}G(t)t^{2k} \, dt = 0 \text{ for } k\ge 1 $$

Since $G$ is even, we can say the same for the integrals from $0$ to $+\infty$, with $2G$ instead of $G$. It remains to change the variable, i.e., to take $$ 2G(t)=2t H(t^2) $$ on $(0,+\infty)$ (since $G$ vanishes near the origin, that will result in $H\in S$ still). The function $H$ has exactly the properties you wanted.

Edit: By Alexei's request, here is the proof that $PQ$ is contracting. We have $\|PQ\|^2=\|PQP\|$. The operator $PQP$ has the kernel $\chi_I(x)\chi_I(x')\int_Ie^{2\pi i(x-x')y}\,dy$, which is supported on $I\times I$ and is bounded by $I$ in absolute value. Thus, by the HS bound, or by the Schur test, we have $\|PQP\|\le |I|^2$, i.e., $\|PQ\|\le |I|$.

It is worth noting that it is actually contracting for an interval $I$ of arbitrary length or even for a set $I$ of finite Lebesgue measure, but the proofs of those facts are not so simple.

The answer is "Yes" though the construction (albeit very standard) is slightly non-obvious.

First, we'll notice that for $f\in L^2$, you can make $f$ and $\hat f$ anything you want on $I=[-0.1,0.1]$, say. That is just because the product $PQ$ of the projections $Pf=f\chi_I$ and $Qf=(\hat f\chi_I)^\vee$ is a contraction in $L^2$, so to solve $Pf=Pg, Qf=Qh$, you just put $f=(g+h)-(Ph+Qg)+(QPh+PQg)-(PQPh+QPQg)+\dots$. Notice that if $g,h$ are real even, the series will consist of real even functions as well, so $f$ will be real even too.

Now just choose some $C^\infty$ even real mollifier $\psi$ with support $(-0.01,0.01)$, $\int_{\mathbb R}\psi=1$, and positive $\hat\psi$. Solve the problem $f|I=0, \hat f|I=\frac{1}{\hat\psi}$ in $L^2$.

Take $F=f*\psi$. It will remain $0$ on $J=\frac 12I$ but we shall get $\hat F=\hat f\hat\psi|I=1$. and $\hat F$ will decay fast. Now convolve with $\psi$ on the Fourier side. You'll get a function $G\in S$ such that $G$ is real even, $G=0$ on $J$ and $\hat G=1$ on $J$. In particular, $$ \int_{\mathbb R}G(t)=1,\quad \int_{\mathbb R}G(t)t^{2k}=0\text{ for }k\ge 1 $$

Since $G$ is even, we can say the same for the integrals from $0$ to $+\infty$, with $2G$ instead of $G$. It remains to change the variable, i.e., to take $$ 2G(t)=2t H(t^2) $$ on $(0,+\infty)$ (since $G$ vanishes near the origin, that will result in $H\in S$ still). The function $H$ has exactly the properties you wanted.

The answer is "Yes" though the construction (albeit very standard) is slightly non-obvious.

First, we'll notice that for $f\in L^2$, you can make $f$ and $\hat f$ anything you want on $I=[-0.1,0.1]$, say. That is just because the product $PQ$ of the projections $Pf=f\chi_I$ and $Qf=(\hat f\chi_I)^\vee$ is a contraction in $L^2$, so to solve $Pf=Pg, Qf=Qh$, you just put $f=(g+h)-(Ph+Qg)+(QPh+PQg)-(PQPh+QPQg)+\dots$. Notice that if $g,h$ are real even, the series will consist of real even functions as well, so $f$ will be real even too.

Now just choose some $C^\infty$ even real mollifier $\psi$ with support $(-0.01,0.01)$, $\int_{\mathbb R}\psi=1$, and positive $\hat\psi$. Solve the problem $f|I=0, \hat f|I=\frac{1}{\hat\psi}$ in $L^2$.

Take $F=f*\psi$. It will remain $0$ on $J=\frac 12I$ but we shall get $\hat F_I=\hat f\hat\psi|I=1$. and $\hat F$ will decay fast. Now convolve with $\psi$ on the Fourier side. You'll get a function $G\in S$ such that $G$ is real even, $G=0$ on $J$ and $\hat G=1$ on $J$. In particular, $$ \int_{\mathbb R}G(t)=1,\quad \int_{\mathbb R}G(t)t^{2k}=0\text{ for }k\ge 1 $$

Since $G$ is even, we can say the same for the integrals from $0$ to $+\infty$, with $2G$ instead of $G$. It remains to change the variable, i.e., to take $$ 2G(t)=2t H(t^2) $$ on $(0,+\infty)$ (since $G$ vanishes near the origin, that will result in $H\in S$ still). The function $H$ has exactly the properties you wanted.

Edit: By Alexei's request, here is the proof that $PQ$ is contracting. We have $\|PQ\|^2=\|PQP\|$. The operator $PQP$ has the kernel $\chi_I(x)\chi_I(x')\int_Ie^{2\pi i(x-x')y}\,dy$, which is supported on $I\times I$ and is bounded by $I$ in absolute value. Thus, by the HS bound, or by the Schur test, we have $\|PQP\|\le |I|^2$, i.e., $\|PQ\|\le |I|$.

It is worth noting that it is actually contracting for an interval $I$ of arbitrary length or even for a set $I$ of finite Lebesgue measure, but the proofs of those facts are not so simple.