I think the space $W$ should be defined a partial order $\leq$ and zero element $0$ firstly and satisfy:
- If $a\leq b$ then $ca\leq cb$, $\forall a,b\in W$ and $0\leq c\in W$.
- If $a\leq b$ then $a-b\leq 0$.
Secondly, a multiply operator $\cdot$ should be defined in $W$ and satisfy $0\leq a\cdot a\stackrel{\triangle}{=}a^2$ for $\forall a\in W$. Also, the inverse operator of $\cdot$ should be defined in $W$ (Alternatively, the inverse element is defined in $W$). That is, if $ab=c$ then $c\stackrel{\triangle}{=}a/b$ for $\forall a,b,c\in W$ and $b\neq 0$ where $/$is the inverse operator of $\cdot$. What is more, these operators should be closed in $W$. Say, if $\forall a,b\in W$ then $a\cdot b\in W$ and $a/b\in W$ if $b\neq 0$. Finally, the operators $\cdot$ and /should satisfy commutative law.
Thirdly, there should have a multiply operator between the elements from $W$ and $V$ because we will define inner product by using this operator.
What is more, to hold the Cauchy-Schwarz inequality, the properties of inner product is important. I believe the Cauchy-Schwarz inequality is valid in a space which define a inner product whose definition is classical. In another word, if a space $V$ have been defined a inner product $(*,*)$ (say, a bilinear form that $V\times V\rightarrow W$) satisfy the following conditions:
- Commutative: $(x,y)=(y,x)$, $\forall x,y\in V$ (If V is a complex space, the right hand side should be dual. But for the sake of simplicity, we ignore it here.)
- Linearity: $(\alpha x+\beta y, z)=\alpha(x,z)+\beta(y,z)$, $\forall x,y,z\in V$ and $\alpha,\beta\in W$.
- Positive define: $(x,x)\geq0$, $\forall x\in V$. The equal sign is valid iff $x=0$ is valid where $0$ donate the zero element in $W$.
Then, by this definition, the Cauchy-Schwarz inequality is valid. The proof are as follow:
For $\forall\lambda\in W$ and $\forall x,y \in V$, we have:
\begin{equation}
0\leq (x+\lambda y,x+\lambda y)=(x,x)+2\lambda(x,y)+\lambda^2(y,y)
\end{equation}
If $y=0$, that is a trivial case and Cauchy-Schwarz inequality is valid obviously. If $y\neq 0$, let $\lambda=-(x,y)/(y,y)$ then we have:
\begin{equation}
0\leq(x,x)-2(x,y)^2/(y,y)+(x,y)^2/(y,y)^2(y,y)\\
(x,y)^2\leq (x,x)(y,y)
\end{equation}
This is the Cauchy-Schwarz inequality.
In fact, Cauchy-Schwarz inequality imply that the inner product of two elements is less than the their product of length because there is an angle between them. And $W$ is a space to measure the inner product of $V$. So I think the conditions I assume at start is reasonable.
Update
I let the solution stand as is, even though it uses non-allowed tools. Maybe someone in the future will find it useful.
Old solution
Here is a solution using the Hölder inequality. If that one is not allowed I suggest that you specify more clearly what is and what is not allowed to use among the "standard tools".
We want to show that (here $w_1$ and $w_2$ are non-negative numbers with $w_1+w_2=1$ and $x$ and $y$ are real numbers)
$$
|w_1x+w_2y|^p\leq w_1|x|^p+w_2|y|^p,
$$
that is
$$
|w_1x+w_2y|\leq \bigl(w_1|x|^p+w_2|y|^p\bigr)^{1/p}
$$
(If any of the involved quantities is $0$ we just note that the inequality is true).
We let $q$ be the dual exponent, i.e. defined such that $1/p+1/q=1$. Then, we have, using first the triangle inequality and then the Hölder inequality,
$$
\begin{aligned}
|w_1x+w_2y|&\leq w_1|x|+w_2|y|\\
&=\bigl(w_1^{1/p}|x|\bigr)(w_1)^{1/q}+\bigl(w_2^{1/p}|y|\bigr)(w_2)^{1/q}\\
&\leq\bigl(w_1|x|^p+w_2|y|^p\bigr)^{1/p}\bigl(w_1+w_2\bigr)^{1/q}\\
&=\bigl(w_1|x|^p+w_2|y|^p\bigr)^{1/p}.
\end{aligned}
$$
Since requested, by Hölder inequality, I mean
$$
a_1b_1+a_2b_2\leq(a_1^p+a_2^p)^{1/p}(b_1^q+b_2^q)^{1/q},
$$
where
$$
a_1=w_1^{1/p}|x|,\quad a_2=w_2^{1/p}|y|,\quad b_1=w_1^{1/q},\quad\text{and}\quad b_2=w_2^{1/q}.
$$
Another for a request about obtaining strict convexity.
If $x$ and $y$ have different signs, then we have strict inequality in the triangle inequality. Thus, we can assume that $x$ and $y$ have the same sign.
Further, one has equality in the Hölder inequality precisely when there exists a constant $c$ such that
$$
|b_1|=c|a_1|^{p-1},\quad\text{and}\quad |b_2|=c|a_2|^{p-1}.
$$
In our case this simplifies to
$$
w_1^{1/q}=cw_1^{(p-1)/p}|x|^{p-1},\quad\text{and}\quad w_2^{1/q}=cw_2^{(p-1)/p}|y|^{p-1}.
$$
Since $1/q=(p-1)/p$ this simplifies to
$$
1=c|x|^{p-1},\quad\text{and}\quad 1=c|y|^{p-1},
$$
and hence to $|x|=|y|$. Since we assumed that $x$ and $y$ had the same sign, we conclude that we have equality precisely when $x=y$. All this implies that we have strict convexity.
Best Answer
Let $f(x)=\sqrt{|x|^{n+\epsilon} \int_0^\infty |\Psi_t(x-y)-\Psi_t(x)|^2 \frac{dt}{t}}$ and $g(x)=\frac{1}{\sqrt{|x|^{n+\epsilon}}}$.
Then, by C-S you have $$\int_{|x|>2|y|} \sqrt{\int_0^\infty |\Psi_t(x-y)-\Psi_t(x)|^2 \frac{dt}{t}} dx \leq \sqrt{ \int_{|x|>2|y|}\frac{1}{|x|^{n+\epsilon}}dx } \sqrt{ \int_{|x|>2|y|} |x|^{n+\epsilon} \int_0^\infty |\Psi_t(x-y)-\Psi_t(x)|^2 \frac{dt}{t}dx}$$
Therefore, your problem boils down to showing that $$ \int_{|x|>2|y|}\frac{1}{|x|^{n+\epsilon}}dx \leq c_n^2 |y|^{-\epsilon} $$ which should follow immediatelly from the convergence of the integral on the LHS, with the proper choice of $c_n$.