I'm looking for the proof (or at least hints how to prove it) of theorem:
Let $H$ be a real Hilbert space and let $f:H\to (-\infty,+\infty]$ be a convex function, bounded from above in some neighbourhood of $x\in H$. Then $f$ is continuous in that point.
Thanks in advance, K.
Let $B\subset H$ be a ball centered at $x$ in which $f(x)<M<\infty$. Without loss of generality we can assume that the radius of $B$ is greater than $1$. (Otherwise we just pass to an equivalent norm on $H$.) Now if $||x-y||<1$, then both points $c=x+(y-x)/||y-x||$ and $d=y+(x-y)/||y-x||$ lie in $B$. Note that $y=(1-||y-x||)x+||y-x||c$ and $x=(1-||y-x||)y+||y-x||d$. Now, by convexity, from the first relation we have $$ f(y)\le (1-||y-x||)f(x)+||y-x||f(c),\qquad\text{or}\\ f(y)-f(x)\le ||y-x||(f(c)-f(x))\le (M-f(x))||y-x||. $$ From the second relation, we obtain $$ f(x)\le (1-||y-x||)f(y)+||y-x||f(d),\qquad\text{or}\\ (1-||y-x||)f(x)+||y-x||f(x)\le (1-||y-x||)f(y)+||y-x||f(d),\qquad\text{or}\\ f(x)-f(y)\le ||y-x||\frac{f(d)-f(x)}{1-||y-x||}\le \frac{M-f(x)}{1-||y-x||}||y-x||. $$ Now we see that $f$ is continuous at $x$.