Prove that there exists $c>0$ such that $\operatorname{supp}(f)\cap K \subset \{x \in K:|f(x)| \geq c\}+\{x \in \mathbb{R}^{n}:|x|<\varepsilon\}.$ Let $f \in C^k(\Omega)$ and $K$ be a compact subset of $\Omega \subset \mathbb{R}^{n}$ and consider $\varepsilon>0$. I'm trying to prove that there exists $c>0$ such that
$$\operatorname{supp}(f)\cap K \subset \{x \in K:|f(x)| \geq c\}+\{x \in \mathbb{R}^{n}:|x|<\varepsilon\}.$$
I think we have to use the fact that $f$ is continuous and thus has a minimum point in compact $K$. But I don't know exactly how.
This is a afirmation from Trèves book, Topological Vector Spaces, Distributions and Kernels.

 A: The assertion is not quite true. There is one case in which it is wrong, and that is when $K$ contains boundary points of $\operatorname{supp} (f)$ but no points with $f(x) \neq 0$ near such a boundary point. The extreme case of that is when $K = \{ x \in \Omega : f(x) = 0\}$. Then $K \cap \operatorname{supp} (f) = \partial K \neq \varnothing$, but
$$\{ x \in K : \lvert f(x)\rvert \geqslant c\} + \{ x \in \mathbb{R}^n : \lvert x\rvert \leqslant \varepsilon\} = \varnothing + \{ x \in \mathbb{R}^n : \lvert x\rvert \leqslant \varepsilon\} = \varnothing$$
for all $c > 0$.
Such things often happen when one only gives an outline of the arguments and avoids writing out all the cumbersome nitty-gritty details. One thinks of the typical situation and overlooks possible pathologies. Nevertheless, at the bottom of this answer I will only give an outline of how I think the proof can be corrected. I believe I haven't overlooked anything, and the missing details can be correctly filled in, but I'm not inclined to fill them in myself. (I don't know of a place where the part of the theorem for which Trèves' proof is incorrect is needed. If that part turns out important too, as the first part is, that would be a reason to dive into the details.)
Probably this is not a serious problem and the proof works if instead of $K \cap \operatorname{supp} (f)$ we consider
$$\operatorname{supp} (f\lvert_K) = \overline{\{ x \in K : f(x) \neq 0\}}\,,$$
but I didn't find the passage in question on a quick leafing through the book, so I can't be sure of that.
Unfortunately, the proof cannot be fixed that easily. But I'll keep the argument for $\operatorname{supp}(f\lvert_K)$ anyway.
Let us work with $\operatorname{supp} (f\lvert_K)$ instead of $K \cap \operatorname{supp} (f)$. For every $c > 0$, consider the set
$$U_c := \{ x \in K : \lvert f(x)\rvert \geqslant c\} + \{ x \in \mathbb{R}^n : \lvert x\rvert < \varepsilon\}\,.$$
Since the second summand is open, $U_c$ is an open set, and it is obviously contained in the analogous set where the inequality in the second summand is non-strict too. Thus it suffices to show that there is a $c > 0$ with $\operatorname{supp} (f\lvert_K) \subset U_c$.
I claim that the family $\{U_c : c > 0\}$ is an open cover of $\operatorname{supp} (f\lvert_K)$. If we accept that claim for the moment, the proof is quickly finished. Since $\operatorname{supp} (f\lvert_K)$ is compact, this cover has a finite subcover, i.e. there is a finite set $\{ c_k : 1 \leqslant k \leqslant m\}$ of (strictly) positive numbers with
$$\operatorname{supp} (f\lvert_K) \subset \bigcup_{k = 1}^m U_{c_k}\,.$$
From the definition of the $U_c$ it is immediate that $c < c' \implies U_{c'} \subset U_c$, and hence we have
$$\operatorname{supp} (f\lvert_K) \subset U_{c_0}$$
for $c_0 = \min\:\{ c_k : 1 \leqslant k \leqslant m\} > 0$.
It remains to prove the claim. Let $x_0 \in \operatorname{supp} (f\lvert_K)$. Then
$$\gamma := \max \{ \lvert f(x)\rvert : x \in K, \lvert x - x_0\rvert \leqslant \varepsilon/2\} > 0$$
since $\{x \in K : \lvert x - x_0\rvert \leqslant \varepsilon/2\}$ is a $K$-neighbourhood of $x_0$, so by definition of the support $f$ cannot vanish identically there. But then clearly $x_0 \in U_{\gamma}$.

Before coming to the fix, I'll note another mistake in Trèves' proof. Directly before the marked part, Trèves wrote

If a compact set $K$ is contained in some open set $\Omega_{j-1}$, we know that $K\cap \operatorname{supp} f_j$ is contained in the neighborhood of order $1/j$ of $K \cap \operatorname{supp}(h_j f) = K \cap \operatorname{supp} f$.

This isn't necessarily correct. The construction only guarantees that $K \cap \operatorname{supp}(f_j)$ is contained in the intersection of $K$ with the $1/j$-neighbourhood of $\operatorname{supp}(f)$. It can be that $K$ lies outside the support of $f$ but intersects the support of $f_j$.
This one however is easily fixable. Let $\varepsilon > 0$. If $K \subset \Omega_m$, then for all sufficiently large $j$ the assertion that $K \cap \operatorname{supp} (f_j)$ is contained in the $\varepsilon$-neighbourhood of $K \cap \operatorname{supp} (f)$ holds. It need not hold for $j = m+1$ (or $j = m + r$ for any fixed $r$), but from some point on it holds. For, let
$$M_{\varepsilon} = \{ x \in K : \operatorname{dist}(x, K \cap \operatorname{supp}(f)) \geqslant \varepsilon\}\,.$$
Then $M_{\varepsilon}$ is compact, and every $x \in \varepsilon$ has a neighbourhood (relative to $\mathbb{R}^n$) on which $f$ vanishes identically - otherwise $x$ would lie in $\operatorname{supp}(f)$. Let
$$r = \min_{x \in M_{\varepsilon}} \operatorname{dist}\bigl(x, \operatorname{supp} (f)\bigr)\,.$$
Then $M_{\varepsilon} \cap \operatorname{supp}(f_j) = \varnothing$, i.e. $K \cap \operatorname{supp}(f_j)$ is contained in the $\varepsilon$-neighbourhood of $K \cap \operatorname{supp}(f)$, for $j > 1/r$.
The theorem to be proved is

Theorem $15.3$. Let $0 \leqslant k \leqslant +\infty$, $\Omega$, be an open set of $\mathbb{R}^n$. Any function $f \in \mathscr{C}^k(\Omega)$ is the limit of a sequence $\{f_j\}$ $(j = 1,2,\dotsc$) of $\mathscr{C}^{\infty}$ functions with compact support in $\Omega$ such that, for each compact subset $K$ of $\Omega$, the set $K \cap \operatorname{supp} f_j$ converges to $K \cap \operatorname{supp} f$.

The convergence of sets is in the sense of the Hausdorff distance here.
The proof begins with taking a normal exhaustion of $\Omega$, that is, a sequence $\Omega_0, \Omega_1, \dotsc$ of open subsets of $\Omega$ such that each $\Omega_j$ is relatively compact in $\Omega_{j+1}$, and $\Omega$ is the union of the $\Omega_j$. Then smooth ($\mathscr{C}^{\infty}$) cut-off functions $h_j$ are constructed, such that $h_j$ is identically $1$ on $\Omega_{j-1}$ and $\operatorname{supp} h_j \subset \Omega_j$.
As a first step, the functions $h_j f$ are considered. These form a sequence in $\mathscr{C}_c^k(\Omega)$ that converges to $f$ in $\mathscr{C}^k(\Omega)$.
In the case $k = +\infty$ we can take $f_j = h_j f$. Then we have $K \cap \operatorname{supp}(f_j) = K\cap \operatorname{supp}(f)$ if $K \subset \Omega_{j-1}$, hence also the second part of the theorem's assertion.
For $k < +\infty$, $h_j f$ is generically only $\mathscr{C}^k$, hence we must approximate that by a $\mathscr{C}_c^{\infty}$ function. That is of course done by convolving $h_j f$ with $\rho_{\delta_j}$ for sufficiently small $\delta_j$. One picks $\delta_j < 1/j$, so that the support of $\varphi_j := \rho_{\delta_j} \ast (h_j f)$ is contained in the $1/j$-neighbourhood of the support of $h_j f$, and additionally $\delta_j$ must be small enough that all of the partial derivatives of $\varphi_j$ of order $\leqslant k$ are within $1/j$ of the corresponding partial derivative of $h_j f$ on all of $\mathbb{R}^n$, and $\operatorname{supp} (\varphi_j) \subset \Omega$.
Then it is clear that $\{\varphi_j\}$ is a sequence in $\mathscr{C}_c^{\infty}(\Omega)$ that converges to $f$ in $\mathscr{C}^k(\Omega)$. Trèves takes $f_j = \varphi_j$, but due to the issues pointed out above I do not see a way to prove the convergence property of the supports without modifying $\varphi_j$ slightly.
Things work out nicely if we have $\operatorname{supp} (h_j f) \subset \operatorname{supp} (\varphi_j)$. Then we always have $K \cap \operatorname{supp}(h_j f)$ contained in the $K \cap \operatorname{supp} (\varphi_j)$, hence a fortiori in every $\varepsilon$-neighbourhood of the latter. And by the argument above, for all sufficiently large $j$ we also have $K \cap \operatorname{supp}(\varphi_j)$ contained in the $\varepsilon$-neighbourhood of $K \cap \operatorname{supp} (h_j f) = K \cap \operatorname{supp}(f)$, thus convergence in the sense of the Hausdorff distance as desired.
But I don't know how to prove $\operatorname{supp}(h_j f) \subset \operatorname{supp}(\varphi_j)$. I don't see how it could be possible that the convolution vanishes identically on an open subset of $\operatorname{supp}(h_j f)$, but I cannot prove that it is impossible. Hence I modify $\varphi_j$ in such a way that the modified function $f_j$ is close enough to $\varphi_j$ that we still have convergence to $f$, and that $\operatorname{supp}(h_j f) \subset \operatorname{supp} (f_j)$. To that end, let $U$ be the interior of the set
$$\{ x \in \operatorname{supp}(h_j f) : \varphi_j(x) = 0\}\,.$$
Then take a function $\eta \in \mathscr{C}_c^{\infty}(\Omega)$ with $U = \{x \in \mathbb{R}^n : \eta(x) \neq 0\}$. The construction is similar to that of a partition of unity on $U$, but one needs to multiply the bump functions $\eta_m$ with a sufficiently fast decreasing coefficient sequence to have the series $\sum c_m \eta_m$ converge in $\mathscr{C}_c^{\infty}$. Multiplying $\eta$ with a small positive constant we can assume that all partial derivatives of order $\leqslant k$ of $\eta$ have modulus $\leqslant 1/j$ everywhere. Then put $f_j = \varphi_j + \eta$.
We have $\operatorname{supp} (f_j) = \operatorname{supp}(\varphi_j) \cup \operatorname{supp} (h_j f)$, and all partial derivatives of order $\leqslant k$ differ from the corresponding derivative of $h_j f$ by no more than $2/j$. Thus $f_j \to f$ in $\mathscr{C}^k(\Omega)$, and we have the desired convergence property for the supports.
