Let $H$ be a separable Hilbert space and $f : [0,1] \to H$ be a $H-$valued mapping.
Suppose that $f$ is weakly continuous in the sense that $\langle f(t), v \rangle_H$ is continuous for any $v \in H$.
Also, suppose that $f$ is norm-decreasing in the sense that $\lVert f(t_2) \rVert_H \leq \lVert f(t_1) \rVert_H$ if $0 \leq t_1 \leq t_2 \leq 1$.
Then, I vaguely remember that $f$ is right-continuous with respect to the norm of $H$.
Is this true? I cannot remmember correctly..
Let $t\in [0,1)$ and fix $\varepsilon_0>0$ such that $t + \varepsilon_0 < 1$.
For $\varepsilon \in (0, \varepsilon_0)$ $$ \begin{align} 0&\le \|f(t)-f(t+\varepsilon)\|^2 = \langle f(t)-f(t+\varepsilon), f(t)-f(t+\varepsilon)\rangle\\ &=\langle f(t),f(t)\rangle -\langle f(t), f(t+\varepsilon)\rangle - \langle f(t+\varepsilon), f(t)\rangle + \langle f(t+\varepsilon), f(t+\varepsilon) \rangle \quad (1) \end{align} $$ Let $$h(\varepsilon) = \langle f(t),f(t)\rangle -\langle f(t), f(t+\varepsilon)\rangle - \langle f(t+\varepsilon), f(t)\rangle$$ By weak continuity $$\lim_{\varepsilon\searrow 0} h(\varepsilon) = \|f(t)\|^2 - \|f(t)\|^2 - \|f(t)\|^2 = -\|f(t)\|^2$$ Now let $$k(\varepsilon) = \|f(t)\|^2 - \|f(t+\varepsilon)\|^2$$ By the monotonicity assumption, $k(\varepsilon)$ is a non-decreasing function of $\varepsilon \in (0, \varepsilon_0)$. Let $U=\inf_{0 < \varepsilon < \varepsilon_0} k(\varepsilon)$. Then $U\ge 0$ and $\lim_{\varepsilon\searrow 0} k(\varepsilon) = U$.
We have $\|f(t+\varepsilon)\|^2 = \|f(t)\|^2 - k(\varepsilon)$ for $\varepsilon\in (0,\varepsilon_0)$. So from $(1)$ we get $$0\le \|f(t) - f(t+\varepsilon)\|^2= h(\varepsilon) + \|f(t)\|^2 - k(\varepsilon)$$ But as $\varepsilon\searrow 0$, $h(\varepsilon)\to -\|f(t)\|^2$, $k(\varepsilon) \ge U \ge 0$ and $k(\varepsilon)\to U$. It follows that $U = 0$ and that $\|f(t) - f(t+\varepsilon)\|\to 0$.