The space of distributions endowed with the topology of uniform convergence on bounded sets is not Fréchet.

I found a state, that the space of distributions on (here:) $\mathbb{R}^{n}$, which is equipped with the topology of uniformly convergence on bounded subsets is not a Fréchet space. As far as i can remember the topology of uniformly convergence is induced by the uniform norm ; which is translation invariant. So to prove that the space is not Fréchet, i have to figure out, that the space is not locally convex or is not a complete metric space. But i have no idea how to prove this.

To clarify your question: By definition $\mathscr D'$ is the dual of the $\mathscr D$ (with its locally convex inductive limit topology) and the topology of uniform convergence on all bounded sets is the locally convex topology on $\mathscr D'$ defined by the semi-norms $p_B(u)=\sup\lbrace |u(\varphi)|: \varphi\in B\rbrace$ for all bounded sets $B$ of $\mathscr D$. This topology is complete (because $\mathscr D$ is a so-called bornological space) and locally convex but it is not metrizable (since there is not countable family of bounded sets such each bounded set is contained in a member of that family, this is so for non-Banach metrisable spaces and hence also in $\mathscr D$).

This is not a separate answer. I only add here the details missing in Jochen's answer. I use here the notation of the post Topology of the Space of Test Functions. Let $\Omega \subseteq \mathbb{R}^n$ be a non-empty open set, and consider the space of test functions $\mathcal{D}(\Omega)$, and let $\mathcal{D'}(\Omega)$ be the set of all continuous linear functionals on $\mathcal{D}(\Omega)$.

Let denote with $\tau'$ the topology of $\mathcal{D'}(\Omega)$ induced by the system of seminorms $p_B(\Lambda)=\sup \{|\Lambda(\phi)|: \phi \in B \}$, where $B$ is any bounded set of $\mathcal{D}(\Omega)$. With this topology, $\mathcal{D'}(\Omega)$ is a Hausdorff locally convex topological vector space (which is called the strong dual of $\mathcal{D}(\Omega)$ in the language of Bourbaki). Moreover, $\mathcal{D'}(\Omega)$ with the topology $\tau'$ is sequentially complete, as you can prove by using the Banach-Steinhaus theorem (see e.g. Rubin, Functional Analysis, Theorem 6.17) or by noting as Jochen did that $\mathcal{D}(\Omega)$ is bornological (see e.g. Bourbaki, Topological Vector Spaces, III, $\S 2$, Proposition 1 (ii') and then use point (iii') of the same Proposition or directly apply Corollary 1 in III, $\S 3$, Section 8).

Now, we show that $\mathcal{D'}(\Omega)$ with the topology $\tau'$ is not metrizable. First of all, not that if $K$ is a compact subset of $\Omega$ with non-empty interior, then by using the property proved in Smooth Functions with Compact Support, we can easily deduce that no neighborhood of $0$ in the space $\mathcal{D}_K$ is bounded. So by a well known result (see e.g. Rubin, Functional Analysis, Theorem (1.39)), $\mathcal{D}_K$ (which is a Fréchet space) is not normable. Then from the result proved in the post Bounded Sets in Fréchet Spaces (see also Bourbaki, Topological Vector Spaces, III, Exercise $\S 1$ Number 14) we conclude that the canonical bornology of $\mathcal{D}_K$ (that is the collection of all bounded sets of $\mathcal{D}_K$) does not have a countable base. Since the topology $\tau_K$ of $\mathcal{D}_K$ coincides with the topology induced on $\mathcal{D}_K$ by the topology $\tau$ of $\mathcal{D}(\Omega)$, we conclude that the canonical bornology of $\mathcal{D}(\Omega)$ has no countable base. Now, assume that $\mathcal{D'}(\Omega)$ with the topology $\tau'$ is metrizable. Then it would have a countable local base at $0$. Since the collection of all finite intersections of sets of the form $$\{ \Lambda \in \mathcal{D'}(\Omega) : p_B(\Lambda) < 1/n \},$$ where $B$ is a bounded set of $\mathcal{D}(\Omega)$ and $n$ is any positive integer, is a local base at $0$, we conclude that there exists a sequence $(B_j)_{j=1}^{\infty}$ of bounded subsets of $\mathcal{D}(\Omega)$ such that the collection of all finite intersections of sets of the form $$W_{B_j,n}= \{ \Lambda \in \mathcal{D'}(\Omega) : p_{B_j}(\Lambda) < 1/n \},$$ where $j, n$ are positive integers, is a local base at $0$. For any positive integer m, let $A_m$ be the closed convex balanced envelope of the bounded set $(m+1)(\cup_{j=1}^{m} B_j) = \{ (m+1) \phi : \phi \in\cup_{j=1}^{m} B_j \}$ . Then every $A_m$ is bounded, since the canonical bornology is adapted (see Bourbaki, Topological Vector Spaces, III, $\S 1$, Section 2). Moreover, since there exists no countable base for the canonical bornology of $\mathcal{D}(\Omega)$, we know there exists a bounded subset $B$ of $\mathcal{D}(\Omega)$ such that $B$ is not contained in $A_m$ for any $m$. So fix a positive integer $m$, and let $\phi_{0} \in B \backslash A_m$. From the Hahn-Banach Theorem (in the form given e.g. in Rudin, Functional Analysis, Theorem 3.7), we know there exists $\Lambda \in \mathcal{D'}(\Omega)$ such that $| \Lambda(\phi_{0})| > 1$ and $| \Lambda(\phi)| \leq 1$ for all $\phi \in A_m$. We conclude that $\Lambda \notin W_{B,1}$, but $\Lambda \in W_{B_1,m} \cap \dots W_{B_m,m}$. So $W_{B_1,m} \cap \dots W_{B_m,m}$ is not contained in $W_{B,1}$. Since for any positive integers $k, j_1,...,j_k$, if we set $m = \max \{ m, j_1,\dots, j_k \}$, we have $$W_{B_1,m} \cap \dots W_{B_m,m,} \subseteq W_{B_1,j_1} \cap \dots W_{B_k,j_k},$$ we conclude that $W_{B_1,j_1} \cap \dots W_{B_k,j_k}$ is not contained in $W_{B,1}$, against the fact that collection of all finite intersections of sets of the form $$W_{B_j,n}= \{ \Lambda \in \mathcal{D'}(\Omega) : p_{B_j}(\Lambda) < 1/n \},$$ where $j, n$ are positive integers, is a local base at $0$. This concludes the proof.

Let us finally note that the argument we have used to prove that $\mathcal{D'}(\Omega)$ is not metrizable since the canonical bornology of $\mathcal{D}$ has no countable base is completely general and it can be used to prove the following general result.

Theorem

Let $E$ be a Hausdorff locally convex (complex or real) topological vector space, and $E'$ the set of continuous linear functionals on $E$. Let $\tau'$ be the topology on $E'$ induced by the family of seminorms $(p_B)$, where $B$ ranges over all the bounded sets of $E$ and $p_B(\Lambda) = \sup \{ |\Lambda(\phi) | : \phi \in B \}$. Then the space $E'$ with this topology (which is called the strong dual of $E$) is metrizable if and only if the canonical bornology of $E$ has a countable base.

Proof. If the canonical bornology of $E$ has a countable base $(B_j)_{j=1}^{\infty}$, then the topology $\tau'$ is induced by the family of seminorms $(p_{B_j})_{j=1}^{\infty}$, and so it is metrizable. The converse follows from the same argument given above.

QED

This theorem is a special case of Bourbaki, Topological Vector Spaces, III, Exercise $\S 3$ Number 2 (c).

PS For the characterization of the metrizability of the weak dual of $E$ (that is $E'$ considered with the weak*-topology), see my post Metrizability of the Weak Dual.