I got a problem with this task. Let $M=\{[f] , f:[0,1] \to \mathbb{F} , \int_0 ^1 |f(x)|^x dx<\infty \}$, where $f$ is a measurable function, be a set and let $\rho(f,g)=\int_0 ^1 |f(x)-g(x)|^x dx$ be a metric. Consider a topology induced by metric $\rho$. Proof, that set $M$ with this topology is or is not locally convex space.
Some hint how on prove it ?
My try - I think it is not locally convex space. Proof:
Let $U\in T(0)$, where $T$ is topology on $M$. We show that $co(U)=M$. Let $\epsilon >0$ such that $B(0,\epsilon)\subset U$ and $n\in \mathbb{N}$ such taht $1/\sqrt{n}<\epsilon$. Consider $f\in M$. Define $I_i=[\frac{i-1}{n},\frac{i}{n}],~i=1,..,n$. Define $f_i=\sqrt{n}f\chi_i,~ i=1,..,n$ where $\chi$ is characteristic function. Then holds $\rho(f_i,0)=\int_0^1|f_i(x)|^xdx=\int_{I_i}|\sqrt{n}f(x)|^xdx\leq \sqrt{n} \int_{I_i}|f(x)|^xdx\leq \frac{1}{\sqrt{n}}<\epsilon$. It will implies $f_i\in B(0,\epsilon)\subset U$ and $f=\sum_1^n \frac{1}{\sqrt{n}}f_i$, so $f\in co(U)$ and then $co(U)=M$. It is correct ? Thanks.
I will prove that $M$ is not locally convex topological vector space.
Consider the open unit ball $B(0,1)$ in $M$, and if possible let $U$ be a convex nbd of $0$ contained in $B(0,1)$. Now there is $\epsilon>0$ such that $U\supseteq B(0,\epsilon)$.
Choose $f\in M$, such that $f(x)=0\ for\ x\in (\frac{1}{2},1]$. Since $f\in M$ we have $\Delta(f):=\int_0^1 |f(x)|^x dx<\infty$.So that we can choose $n\in \Bbb N$ such that $ n^{-1/2} \Delta(f)<\epsilon$. Now by the continuity of indefinite integral of $t\rightarrow \int_0^t|f(x)|^xdx$ we have points $0=t_0<t_1<......<t_n=1$ such that $\int_{t_{i-1}}^{t_i}|f(x)|^xdx=n^{-1}\Delta(f)$ for $1≤i≤n$. Define $f_i=nf\chi_i,~ i=1,..,n$ , where $\chi_i$ is the characteristic function of $[t_{i-1},t_i]$. Therefore $$\rho(f_i,0)=\int_0^1|f_i(x)|^xdx=\int_{t_{i-1}}^{t_i}n^x|f(x)|^xdx≤\int_{t_{i-1}}^{t_i}n^{1/2}|f(x)|^x\ dx =n^{1/2}\frac{\Delta(f)}{n}<\epsilon$$
Now $f=\sum_1^n \frac{1}{n}f_i$ and $f_i\in B(0,\epsilon)\subseteq U$ and since $U$ is convex we have $f\in U$.
What we have showed that for any $f\in M$ with $f(x)=0\ for\ x\in (\frac{1}{2},1]$, $f\in U\subseteq B(0,1)$. But then the functions of the form $$g_n(x)=n^x\ for\ x\in [0,1/2] \ and\ g_n(x)=0 \ for\ x\in (1/2,1]\ and\ n\in \Bbb N$$ are in $U$. But notice that $g_n\in U\subseteq B(0,1)$ for each $n\in \Bbb N$ and $\rho(g_n,0)=\frac{n^{1/2}-1}{log_e n}\rightarrow \infty$, which is impossible. Hence $M$ has no local base $\mathscr B_0$ at $0$ such that members of $\mathscr B_0$ are convex. Hence $M$ cannot be locally convex topological vector space.