Inequality involving $\lambda$-convex functional. I stuck with the following problem:
Let $F : C^1_0[0,1] \rightarrow \mathbb{R}$ be a $\lambda$-convex functional: there exist a $\lambda > 0$, such that
\begin{equation}
F((1-\sigma)u + \sigma v) \leq (1-\sigma) F(u) + \sigma F(v) - \frac{\lambda}{2}\sigma(1-\sigma) \int^{1}_{0}{|u(x) - v(x)|^2 dx}
\end{equation}
for all $u,v \in C^1_0[0,1]$ and $\sigma \in [0,1]$.
I have already proofed the following 2 statements
If $\overline{u} \in C^1_0[0,1]$ is a stationary point of $F$ than:
\begin{equation}
\frac{\lambda}{2}\int^1_{0}{|u(x) - \overline{u}(x)|^2 dx} \leq F(u) - F(\overline{u})
\end{equation}
and $\overline{u}$ is a global minimizer of $F$.
Now I struggle with the following part:
Let $U \subset C^1_0[0,1]$ be convex, such that for all $u \in U$ there exist a gradient $\zeta_u \in L^2(0,1)$; which means that if $F$ is frechet-differentiable in $u$ and $\zeta_u$ is the riesz-representation of the frechet-derivative $DF(u)$ in $L^2(0,1)$:
\begin{equation}
\delta F(u)[\xi] = \int^{1}_{0}{\zeta_u(x) \xi(x) dx}
\end{equation}
for all $\xi \in C^1_0[0,1]$. Let $\overline{u}$ be a stationary point of $F$:
Show that for all $u \in U$:
\begin{equation}
2 \lambda [F(u) - F(\overline{u})] \leq \int^{1}_{0}{\zeta_u(x)^2 dx}
\end{equation}
I tried so far the following:
\begin{align*}
            F((1-\sigma)u + \sigma v) &\leq (1-\sigma)F(u) + \sigma F(v) - \frac{\lambda}{2}\sigma(1-\sigma)\int^{1}_{0}{|u(x) - v(x)|^2 dx} \\
            &\iff  \frac{\lambda}{2}\sigma(1-\sigma)\int^{1}_{0}{|u(x) - v(x)|^2 dx} \leq F(u) + \sigma(F(v) - F(u)) - F(u + \sigma (v - u )) \\
            &= -\left( F(u + \sigma (v - u )) - F(u) - \sigma(F(v) - F(u)) \right) \\
            &\iff \frac{\lambda}{2}\int^{1}_{0}{|u(x) - v(x)|^2 dx} \leq \frac{1}{\sigma} \frac{1}{\sigma - 1}\left( F(u + \sigma (v - u )) - F(u) - \sigma(F(v) - F(u)) \right)
        \end{align*}
which leads through substituting $y = \sigma - 1$ to
\begin{align*}
            \frac{\lambda}{2}\int^{1}_{0}{|u(x) - v(x)|^2 dx} &\leq \frac{1}{y+1}\frac{1}{y}\left( F(u + (1+y) (v - u )) - F(u) - (1+y)(F(v) - F(u)) \right) \\
            &=\frac{1}{y+1}\frac{1}{y}\left( F(v + y(v - u )) - F(v) - y(F(v) - F(u)) \right)
        \end{align*}
Now $DF(u)[\xi] = dF(u)[\xi] = \delta F(u,\xi)$ since frechet-differentiable functions are gateux-differentiable and the first variation matches the frechet-derivative.  From then on I can't figure out how to come to the desired inequality.
I would be thankful for every hint.
 A: I think I proofed it:
\begin{align*}
            F((1-\sigma)u + \sigma v) &\leq (1-\sigma)F(u) + \sigma F(v) - \frac{\lambda}{2}\sigma(1-\sigma)\int^{1}_{0}{|u(x) - v(x)|^2 dx} \\
            &\iff  \frac{\lambda}{2}\sigma(1-\sigma)\int^{1}_{0}{|u(x) - v(x)|^2 dx} \leq F(u) + \sigma(F(v) - F(u)) - F(u + \sigma (v - u )) \\
            &= -\left( F(u + \sigma (v - u )) - F(u) - \sigma(F(v) - F(u)) \right) \\
            &\iff \frac{\lambda}{2}\int^{1}_{0}{|u(x) - v(x)|^2 dx} \leq \frac{1}{\sigma} \frac{1}{\sigma - 1}\left( F(u + \sigma (v - u )) - F(u) - \sigma(F(v) - F(u)) \right) \\
            &= \frac{1}{\sigma} \frac{1}{\sigma - 1}\left( F(u + \sigma (v - u )) - F(u) \right) + \frac{1}{1-\sigma}(F(v) - F(u))
        \end{align*}
for $\sigma \rightarrow 0$ we get
\begin{align*}
        0 &\leq \frac{\lambda}{2}\int^{1}_{0}{|u(x) - v(x)|^2 dx} \\
        &\leq \frac{1}{\sigma} \frac{1}{\sigma - 1}\left( F(u + \sigma (v - u )) - F(u) \right) + \frac{1}{1-\sigma}(F(v) - F(u)) \\
        &\stackrel{\sigma \rightarrow 0}{\rightarrow } -\delta F(u,v-u) + (F(v) - F(u)) \\
        &= \int^{1}_{0}{\zeta_u(x)(u(x)-v(x))dx} + (F(v) - F(u)) 
    \end{align*}
therefore
\begin{equation*}
        F(u) - F(v) \leq \int^{1}_{0}{\zeta_u(x)(u(x)-v(x))dx} 
    \end{equation*}
and with the holder-inequality
\begin{equation*}
        F(u) - F(v) \leq \left(\int^{1}_{0}{\zeta_u(x)^2 dx}  \right)^{\frac{1}{2}} \left( \int^{1}_{0}{|u(x)-v(x)|^2 dx}  \right)^{\frac{1}{2}}
    \end{equation*}
If we take $v = \overline{u} \in U$ and $u = v \in U$, such that  the convex combination in the calculation above $v + \sigma(\overline{u} - v) \in U$ and
$v-u \in U \subset C^1_0[0,1]$, then from $(i)$ and $(ii)$ it follows that
$\overline{u}$ is a global minimizer, therefore $F(v) - F(\overline{u}) \geq 0$ and we get
\begin{equation*}
        F(v) - F(\overline{u}) \leq \left(\int^{1}_{0}{\zeta_v(x)^2 dx}  \right)^{\frac{1}{2}} \left(\frac{2}{\lambda} (F(v) - F(\overline{u})) \right)^{\frac{1}{2}}
    \end{equation*}
therefore
\begin{equation*}
        \left( \frac{\lambda}{2} (F(v) - F(\overline{u})) \right)^{\frac{1}{2}} \leq \left(\int^{1}_{0}{\zeta_v(x)^2 dx}  \right)^{\frac{1}{2}}
    \end{equation*}
also
\begin{equation*}
        \frac{\lambda}{2}(F(v) - F(\overline{u})) \leq \int^{1}_{0}{\zeta_v(x)^2 dx} < \infty
    \end{equation*}
since $\zeta_v \in L^2(0,1)$ the right integral is finite, and since the interval is compact $(v-\overline{u})^2 \in L^2(0,1)$.
Somehow the factor is $\frac{\lambda}{2}$ instead of $2 \lambda$ did I make a mistake in my calcuations?
